
Class ~>5"3^ 

mum* 1 

Book J - 

GopyrightN? 

COPYRIGHT DEPOSIT. 



PLANE TRIGONOMETRY 



BY 



JAMES M. TAYLOR, A.M., LL.D. 

Professor of Mathematics, Colgate University 





1 g 


i • 









BOSTON U.S.A. 
GINN & COMPANY, PUBLISHERS 

1904 



LIBRARY of CONGRESS 
Two Copies Received 

JAN 21 1904 

*^ Copyright Entry 






No. 



:opy 3 



5T33 



Copyright, 1904 
By JAMES M. TAYLOR 



ALL RIGHTS RESERVED 



PREFACE 



This book is designed to meet the needs of beginners who 
wish to master the fundamental principles of Trigonometry. 
The author's aim has been to prepare a text-book which shall 
be clear and practical, jet thoroughly scientific. 

The proofs of formulas are simple but rigorous. The use of 
directed lines is consistent; the directions of such lines in the 
figures are usually indicated by arrowheads, and these lines 
are always read from origin to end. Both trigonometric ratios 
and trigonometric lines are employed, but at first the ratios 
are used exclusively until they have become fixed in mind and 
familiar. It is proved that the ratios are the measures of the 
lines, and that, therefore, any relation which holds true for 
the one holds true for the other also. 

The distinction between identities and equations is empha- 
sized in definition, treatment, and notation. The solution of 
trigonometric equations is scientific and general. The trigo- 
nometric ratios are defined in pairs as reciprocals of each 
other both to aid the memory and to emphasize one of the 
most important of their fundamental relations. In the reduc- 
tion of the functions of any angle to those of an acute angle, 
the theorems concerning the functions of — A and 90° + A are 
made fundamental. The addition formulas are proved for 
positive or negative angles of any quadrant and from them 
are deduced the other formulas concerning the functions of 
two or more angles. When two or more figures are used in 
a proof, the same phraseology always applies to each figure. 



iv PREFACE 

In the first chapter will be found a table of natural functions 
at intervals of 5°, which the student is to verify to two places 
by construction and measurement, and many interesting prob- 
lems have been introduced in order that the subject may be 
made more attractive and profitable to the beginner. The 
angles and numbers in these problems are so chosen that this 
table contains all that is needed for their solution. 

In Chapter VIII complex number is expressed as an arith- 
metic multiple of a quality unit in its trigonometric type 
form, and the fundamental properties of such number are 
demonstrated. The proof of De Moivre's theorem is simple 
and general, and its meaning and use are fully illustrated. 

Before beginning the study of complex number one should 
gain a clear idea of the sum, or resultant, of two directed 
forces, and by repeated experiment make this idea familiar. 

It is believed that the order of the text is the best for begin- 
ners ; but, with the exception of a few articles, Chapter I may 
be omitted by those who are prepared to take up at once the 
general treatment in Chapter II. Too much stress cannot be 
laid on careful and accurate construction and measurement in 
the first chapter. Chapters VII and VIII and the latter parts 
of Chapter VI may be omitted by those who wish a shorter 
course. 

In preparing this book the author has consulted the best 
authorities, both American and European. Many of the exam- 
ples have been taken from these sources. The author takes 
this opportunity to express to many teachers and other friends 
his appreciation of their valuable suggestions and their en- 
couragement in the preparation of this work. 

James M. Taylor. 

Colgate University, December, 1903. 



CONTENTS 



CHAPTER I 

TRIGONOMETRIC RATIOS OF ACUTE ANGLES 

Section Page 

1-3. Trigonometric ratios defined, one-valued 1,2 

4. Construction of angles having given ratios 3, 4 

5, 6. Approximate values, and changes from 0° to 90° ... 5-7 

7-9. Trigonometric ratios of angles of triangle and of co-angles 8, 9 

10,11. Trigonometric ratios of 45°, 30°, 60° 10,11 

12, 13. Solving right triangles 11-13 

14, 15. Definitions and problems 14-21 



CHAPTER II 

TRIGONOMETRIC RATIOS OF POSITIVE AND NEGATIVE ANGLES OF ANY SIZE 

16. Positive and negative angles of any size 22, 23 

17-19. Coterminal angles. Quadrants 24,25 

20, 21. Trigonometric ratios of any angle 26, 27 

22,23. Laws of quality. Sin.- 1 *;, cosr 1 ^ • • • 28-30 

24, 25. Fundamental identities. Proof of identities 31-35 

26. Changes in ratios as angle changes from 0° to 360° . . . 36-38 

27. Trigonometric ratios of 0°, 90°, 180°, 270° 39 

28, 29. Trigonometric ratios of - A and 90° + A 40-43 

30, 31. Trigonometric ratios of 90° - A, 180° - A, n 90° ± A . . 44, 45 

32, 33. Trigonometric lines representing the ratios 46-51 



VI CONTENTS 



CHAPTER III 



TRIGONOMETRIC RATIOS OF TWO ANGLES 

Section Page 

84-37 . Trigonometric ratios of the sum and difference of two angles 52-56 

38, 39. Trigonometric ratios of twice and half an angle .... 57-59 

40. Sums and differences of trigonometric ratios 60-63 



CHAPTER IV 

SOLUTION OF RIGHT TRIANGLES WITH LOGARITHMS 

41-48. Properties and computation of logarithms 64-68 

49. Right-angled triangles 69-71 

50, 51. Isosceles triangles and regular polygons 72, 73 



CHAPTER V 
SOLUTION OF TRIANGLES IN GENERAL 

52. Law of sines and law of cosines 74, 75 

53-55. The four cases solved without logarithms . . . . . . 76, 77 

56. Law of tangents . 78, 79 

57-59. First three cases solved with logarithms 80-87 

60. Trigonometric ratios of half angles and Case (iv) solved 

with logarithms 88-91 

61. Areas of triangles 92 

62-64. Circumscribed, inscribed, escribed circles 93, 94 



CONTENTS vii 



CHAPTER VI 

RADIAN MEASURE, GENERAL VALUES, TRIGONOMETRIC EQUATIONS, 

INVERSE FUNCTIONS 

Section p AGE 

65-67. Radian measure of angles 95-97 

68. Principal values 98 

69-71. Angles having the same trigonometric ratio .... 99, 100 

72, 73. Solution of trigonometric equations ....... 101-104 

74. Trigonometric functions 105 

75, 76. Inverse trigonometric functions 106-109 



CHAPTER VII 

PERIODS, GRAPHS, IMPORTANT LIMITS, COMPUTATION OF TABLE, 
HYPERBOLIC FUNCTIONS 

77. Periods of the trigonometric functions 110 

78-81. Graphs of the trigonometric functions 111-114 

82. Limit of the ratio of sin or tan 6 to 6 115 

83, 84. Computation of trigonometric functions 116, 117 

85,86. Hyperbolic functions 118,119 



CHAPTER VIII 
COMPLEX NUMBERS. DE MOIVRE'S THEOREM 

87. Quality units ±1, ± V^~l 120 

88, 89. Directed lines and forces 121, 122 

90. Complex numbers 123,124 

91, 92. General quality unit. Products of quality units . . 125, 126 

93. De Moivre's theorem. Quotients of quality units . . 127, 128 



viii CONTENTS 

Section Page 

94, 95. Products and quotients of complex numbers .... 128, 129 

96, 97. The gth roots of cos + i sin and (cos + i sin <p) r . 130, 131 

98. Exponential form for cos + i sin 132, 133 



CHAPTER IX 

MISCELLANEOUS EXAMPLES 

Trigonometric identities 134-136 

Trigonometric equations and systems 137-141 

Problems involving triangles 142-146 

Problems involving areas and regular polygons . . . . . . 146-147 

Formulas 148-151 



ANSWERS 153-171 



PLAKE TRIGONOMETRY 

CHAPTEE I 
TRIGONOMETRIC RATIOS OF ACUTE ANGLES 

1. Let A denote the number of degrees in the acute angle 
XOB] then A is the numerical measure, or measure, of this 
angle, and we can write Z XOB = A. 

Prom any point in either side of the angle XOB, as P, 
draw PM perpendicular to the 
other side. 

Observe that is the vertex 
of the angle, and M is the foot of 
the perpendicular drawn from P. 
This lettering should be fixed in 
mind so that in the following defi- 
nitions the lines MP, OM, and OP shall always mean the same 
lines as in fig. 1. 

The six simple ratios (three ratios and their reciprocals) 
which can be formed with the three lines MP, OM, OP are 
called the trigonometric ratios of the angle XOB, or A. 

These ratios are named as follows : 

The ratio MP /OP is the sine of A ; 

and its reciprocal OP /MP is the cosecant of .4. 

The ratio OM/OP is the cosine of A ; 

and its reciprocal OP/OM is the secant of A. 

The ratio MP/OM is the tangent of A ; 

and its reciprocal OM/MP is the cotangent of A. 

1 




2 PLANE TRIGONOMETRY 

For brevity the sine of A is written sin A ; the cosine of A, 
cos A; the tangent of A, tan A; the cotangent of A, cot A; the 
secant of A, sec A ; and the cosecant of A, esc A. 

Observe that sin A is a compound symbol which, taken as a 
whole, denotes a number. The same is true of cos A, tan J., etc. 

Ex. 1. What four trigonometric ratios of the angle A involve the 
line MP? the line OM? the line OP? 

Ex. 2. What trigonometric ratios are reciprocals of each other ? 
Ex. 3. Which is the greater, tan A or sec A ? cot A or esc A ? Why ? 
Ex. 4. Can sin A or cos A exceed 1 ? Why ? 

2. ^4m/ trigonometric ratio of a given angle has only one 
value. 

Let XOB be any acute angle. Draw PM A. OX, P f M' ± OX, 

P"M" _L OB ; then, by § 1, 

MP M'P' 
smXOB = -—> — -7j 
OP OP' 

or M"P"/OP". (1) 

From the similarity of the A 
P>, X 0¥P ? 0¥'P', OM"P" it follows 
that the three ratios in (1) (or 
their reciprocals) are all equal; 
hence sin XOB (or esc XOB) has but one value. 

Also, from the similarity of these A, each of the other trigo- 
nometric ratios of Z XOB has only one value. 

3. Two acute angles are equal if any trigonometric ratio of 
the one is equal to the sam,e ratio of the other. 

Take O x P x in fig. 3 equal to OP in fig. 2, and draw 
P 1 M 1 A. 1 X 1 . We are to prove that 

if sin X 1 O l P 1 = sin XOP, 

i.e. if M X P X J O x P x = MP / OP, (1) 

then Z X x O x P x = Z XOP. (2) 




TKIGONOMETKIC EATIOS OF ACUTE ANGLES 




By construction, 1 P 1 = OP. 
Hence, from (1), M 1 P 1 = MP. 

Therefore, by Geometry, the right tri- 
angles OxPiJ/i and OP M are equal in 
all their parts. O l 

Hence Z X 1 1 P 1 = Z XOP. 

In like manner the student should prove the equality of 
two acute angles when any other trigonometric ratio of the 
one is equal to the same ratio of the other. 

4. Having given the value of any trigonometric ratio of an 
acute angle, to construct the angle and obtain the values of its 
other trigonometric ratios. 

This problem will be illustrated by particular examples. 

Ex. 1. If A is an acute angle and cos A = 3/5, construct A and find 
the values of its other trigonometric ratios. 

Here cos A = OM / OP = 3/5. 
Hence, if OP = 5 units, OM = 3 units. 

Let 0, in fig. 4, be the vertex of the angle 
A, and OX one of its sides. 

On OX, to some scale, lay off OM equal 
to 3 units, and at M draw MS ± OX. 

With as a center and with a radius 
equal to 5 units, draw an arc cutting MS in 
some point as P. Draw OPB. 

Then ZXOB = A. 

For cos XOB = 3 / 5 = cos A. 

Hence, by § 3, Z XOB = A. 

Again, MP = Vo 2 - 3 2 units = 4 units. 

Hence sin A =4/5, cscJL = 5/4; 

cosJ. = 3/5, sec A =5/3; 

tan^l =4/3. cot A = 3/4. 

Observe that 5 is the numerical measure of OP, 4 of MP, and 3 of OM. 




Fig. 4 



M X 



§1 



PLANE TRIGONOMETRY 



Ex. 2. If A is an acute angle and sin ^4 
the values of its other trigonometric ratios. 

Here 



2/3, construct A and find 




M X 



Fig. 5 



sinA = MP/OP = 2/S. 

Hence, if OP = 3 units, 
MP = 2 units. 

At If, in fig. 5, draw MS ± OX and 
lay off MP equal to 2 units. 

With P as a center and 3 units as a 
radius, strike an arc cutting OX in some 
point, as 0. Draw OPB. 

Then ZXOB = A. 

For sin XOB = 2/3 = sin^i. 

Hence, by § 3, Z XOB = A. 
OM = V3 2 - 2 2 units = V5 units, 
sin A = 2/3, esc ^4=3/2; §3 

cos A = V5/3, sec A = 3/V5; 

tan A = 2/V5, cot A = V5/2. 

If the vertex of the angle A were required to be at some fixed point 
on OX, as O, we would draw OK ± OX, lay off OQ equal to 2 units, 
through Q draw QiV parallel to OX, and with as a center and 3 units 
as a radius, strike an arc cutting QN at P ; then draw OPB as before. 

By using a protractor, i.e. a graduated semicircle, we find that Z XOB, 
or A, is an angle of about 42°. 

Note. The student should form the habit of gaining a clear idea of the 
size of an angle from the value of any one of its trigonometric ratios. 



Again, 
Hence 



EXERCISE I 

Construct the acute angle A, obtain the values of all its trigonometric 
ratios, and find its size in degrees, when : 



1. sin^l = 2/5. 

2. sin A = 4/5. 

3. cos^l = 3/4. 

4. cos .4 = 1/3. 

5. tan A = 1/4. 



6. tan A = 4/3. 

7. cot^4 = 5/2. 

8. cot^. = 1/3. 

9. sec A = 5/3. 
10. sec A =4/3. 



11. esc .4 =5/2. 

12. esc A =3/2. 

13. tan .4 = 4, or 4/1. 

14. cot^ =7, or 7/1. 

15. tan A = 9. 



TRIGONOMETRIC RATIOS OF ACUTE ANGLES 



16. Express each of the trigonometric ratios of an acute angle A in 
terms of its sine, writing (sin-4) 2 in the form sin 2 .A. 



In fig. 1, let 

Then 

Whence 

and 
Hence 



OP = 1. 
MP /OP = MP/1, i.e. sin A is the measure of MP. 
MP = sin A, 



sin 2 ^.. 



O'M = V OP"- - MP 2 = VT 
cos A = OM/ OP = Vl - sin 2 ^L; 
. sec A — 1 / Vl — sin 2 .4. 
tan A = MP/ OM = sin A/y/l- sin-M 



§1 



.-. cot A = Vl — sin 2 A /sin A. 
esc A = OP /MP = 1/sin A. 

17. Express each of the trigonometric ratios of an acute angle in terms 
of its cosine. 

In fig. 1, let OP = 1. 

Then OM / OP = 03// 1, i.e. cos A is the measure of OM. 

Whence OM = cos A, 

and J/P = ^OP 2 - OM' 2 = Vl - cos 2 ^i, etc. 

18. Express each of the trigonometric ratios of an acute angle in terms 
of its tangent. 

In fig. 1, let 0M = 1. 

Then MP/ OM = MP/1, i.e. tan A is the measure of MP. 

Whence MP = t&nA, 

and OP = 



■ VW 2 + MP 2 = Vl + tan 2 .l. etc. 



5. To find approximately by measurement the values of tlie 
trigonometric ratios of any given angle. 

Ex. 1. Eind by construction and 
measurement the values of the six trigo- 
nometric ratios of 40°. 

With a protractor lay off A XOB = 40°. 

Take OP any convenient length, say 10 
units (the longer the better), and draw 
PM _L OX. By careful measurement we 
find that 

MP = 6.4 units, 

OM — 7.7 units, approximately. Fig. 6 




PLANE TRIGONOMETRY 



Hence, as approximate values, we have 

sin 40° = 6.4/10 = 0.64, esc 40° = 10/6.4 = 1.56; 

cos 40° = 7.7/10 = 0.77, sec 40° = 10/7.7 = 1.3 ; 

tan40° = 6.4/7.7 = 0.83, cot40° = 7.7/6.4 = 1.2. 

Observe that instead of taking OP equal to 10 units, we could take 
OM equal to 10 units. 

Ex. 2. By construction and measurement find the approximate values 
of the sine, cosine, and tangent of 5°, 10°, 15°, 20°, 25°, 35°, and compare 
the results obtained with their values given in the table below. 

On a scale of 20 to an inch take OM equal to 100 units and draw MP 
± OM. With O as their common vertex and OM as their common lower 
side, draw the angles 5°, 10°, etc. 

Observe that the cosecant, the secant, and the cotangent of any angle 
can be obtained by taking the reciprocal of the sine, the cosine, and the 
tangent respectively of the same angle. 



Angle 


sin 


CSC 


cos 


sec 


tan 


cot 


1° 


0.0175 


57.2987 


0.9998 


1.0002 


0.0175 


57.2900 


5° 


0.0872 


11.4737 


0.9962 


1.0038 


0.0875 


11.4301 


10° 


0.1736 


5.7588 


0.9848 


1.0154 


0.1763 


5.6713 


15° 


0.2588 


3.8637 


0.9659 


1.0353 


0.2679 


3.7321 


20° 


0.3420 


2.9238 


0.9397 


1.0642 


0.3640 


2.7475 


25° 


0.4226 


2.3662 


0.9063 


1.1034 


0.4663 


2.1445 


30° 


0.5000 


2.0000 


0.8660 


1.1547 


0.5774 


1.7321 


35° 


0.5736 


1.7434 


0.8192 


1.2208 


0.7002T 


1.4281 


40° 


0.6428 


1.5557 


0.7660 


1.3054 


0.8391 


1.1918 


45° 


0.7071 


1.4142 


0.7071 


1.4142 


1.0000 


1.0000 


50° 


0.7660 


1.3054 


0.6428 


1.5557 


1.1918 


0.8391 


55° 


0.8192 


1.2208 


0.5736 


1.7434 


1.4281 


0.7002 


60° 


0.8660 


1.1547 


0.5000 


2.0000 


1.7321 


• 0.5774 


65° 


0.9063 


1.1034 


0.4226 


2.3662 


2.1445 


0.4663 


70° 


0.9397 


1.0642 


0.3420 


2.9238 


2.7475 


0!3640 


75° 


0.9659 


1.0353 


0.2588 


3.8637 


3.7321 


0.2679 


80° 


0.9848 


1.0154 


0.1736 


5.7588 


5.6713 


0.1763 


85° 


0.9962 


1.0038 


0.0872 


11.4737 


11.4301 


0.0875 


89° 


0.9998 


1.0002 


0.0175 


57.2987 


57.2900 


0.0175 



In the above table observe how each trigonometric ratio of A changes 
as A increases from 1° to 89°. 



TRIGONOMETRIC RATIOS OF ACUTE ANGLES 




6. Changes of the trigonometric ratios of the angle A as A 
increases from 0° to 90°. 

In fig. 7 conceive the line OP to revolve from OX to OB, i.e. 
suppose Z.XOP, or A, to increase 
from 0° to 90°. 

When OP coincides with OX, 

MP = 0, and OM = OP. 

When OP coincides with OB, 

MP = OP, and OM = 0. 

1, 



Hence sin 0° = 0, sin 90° = 
cos 0° = 1, and cos 90° = 0. 

Therefore when A increases 
from 0° to 90° sin A increases from to 1 and cos A decreases 
from 1 to 0. 

Example. What is the value of tan 0° ? sec 0° ? cot 90° ? esc 90° ? 

When OP is very near to OB, OM is very small ; hence 
MP J OM is very large, and the nearer OP is to OB, the larger 
is MP J OM, or tan A. So that as A approaches 90°, tan A 
increases rapidly and can exceed any constant number how- 
ever great ; that is, tan A becomes an infinite (denoted by oo). 
Hence as A increases from 0° to 90°, tan A increases from 
to oo ; likewise sec A increases from 1 to oo. 

Similarly, when OP approaches nearer and nearer to OX, 
OM J MP, or cot A, and OP / MP, or esc A, become infinites. 

Hence as the angle A increases from 0° to 90°, 

sin A increases from to 1, 
cos A decreases from 1 to 0, 
tan A increases from to oo, 
cot A decreases from oo to 0, 
sec A increases from 1 to oo, 
esc A decreases from oo to 1. 



PLANE TRIGONOMETRY 




B 



7. Trigonometric ratios of the acute angles of a right triangle. 

Let A and B be the measures of 
the acute angles of the right tri- 
angle ABC, a the measure of the 
« side opposite the angle A, b that of 
the side opposite B, and c that of 
J C the hypotenuse; then, by § 1, we 
have 



Fig. 8 



sin A = - ■■ 
c 



cos A 



tan A 



side opposite 

hypotenuse 

b _ side adjacent 

c hypotenuse 

_ a _ side opposite 

6 side adjacent 

Similarly we have 

sin B = b/c, 
cos B = a/c, 
tan I? = b/a, 

8. Complementary angles 



esc A 



sec A 



cot ^4 = - • 
a 



esc B = c/b ; 
sec B = c/a; 
cot B = a/b. 

Two angles are said to be com- 



plementary when their sum is 90°. 

E.g., the complement of 35° is 90° — 35°, or 55°; the complement of 
70° is 90° - 70°, or 20° ; the complement of A is 90° - A ; and in any 
right triangle, as ABC fig. 8, the acute angles A and B are complementary 
angles. 

9. Trigonometric ratios of complementary angles. If Z CAB 

= A, then Z CBA = 90° - A. Hence, by § 7, we have 

sin (90° — A) = b/c = cos A, 
cos (90° — A) = a/c == sin A, 
tan (90° — A) = b/a= cot .4, 
cot (90° -A) = a/b = tan ^4, 
sec (90° — A) = c/a = esc ^4, 
esc (90° - 4) = c/ft t= sec J. 



TRIGONOMETRIC RATIOS OF ACUTE ANGLES 




Fig. 9 



If we call the cosine the co-ratio of the sine, the sine the 
co-ratio of the cosine, the cotangent the co-ratio of the tangent, 
the tangent the co-ratio of the co- 
tangent, etc., then the six identities 
above can be summed up as follows : 

Any trigonometric ratio of an acute 
angle is equal to the co-ratio of its 
complement. 

E.g., since 60° and 30° are comple- 
mentary angles, we have 

sin 60° = cos 30°, tan 60° = cot 30°, esc 60° = sec 30°. 
Again, since 45° is the complement of itself, we have 

sin 45° = cos 45°, tan 45° = cot 45°, esc 45° = sec 45°. 

Ex. 1. From the upper half of the table in § 5 obtain the lower half. 

Since 50° + 40° = 90°, sin 50° = cos 40° = 0. 7660 ; 

since 55° + 35° = 90°, sin 55° = cos 35° = 0.8192 ; 

since 60° + 30° = 90°, sin 60° = cos 30° = 0.8660 ; etc. 

Ex. 2. The angle A being acute, find the value of A in the equation 
sin A = cos 2 A. (1) 

If in equation (1) we substitute for cos 2 A its * identical expression 
sin (90° — 2-4), by Algebra we obtain the equivalent equation (2). 

* Algebraic definitions. Two numeral expressions which denote the same 
number, or any two mathematical expressions which denote equal numbers 
for all values of their letters, are called identical expressions. E.g., the numeral 
expressions 4x3 and 8 + 4 are identical, so also are the literal expressions* 
(a + b) (a - b) and a 2 - b' 2 , or cos A and sin (90° - A). 

An equality is the statement that two mathematical expressions denote the 
same number. 

An equality whose members are identical expressions is called an identity. 
An identity is to be proved. 

An equality whose members are not identical expressions is called an 
equation. An equation is to be solved. 

The sign of identity, = read " is identical with," is often used instead of 
the sign of equality = in writing an identity whose members involve one or 
more letters. E.g., to indicate that the equality sin J. = cos (90° — A) is an 
identity and not an equation we write sin A = cos (90° — A) . 

Since we know that any equality which involves only numerals must be an 
identity, the sign of identity is used only in writing literal identities. 



10 



PLANE TRIGONOMETRY 



From (2), by § 3, 



sin A 
A 



sin (90° -2 A). 
90° -2 A, ..A 



(2) 



30°. 



Equation (1) is a trigonometric equation, and the only value of the 
acute angle A which will satisfy it is 30°. 



EXERCISE II 

1. By § 9, cos 30° equals what ? sin 60° ? cot 35° ? tan 15° ? sec 85°? 
esc 76° ? sin 73° 14' ? cos 65° 43'? 

A being an acute angle, find its value in each of the following equations: 

2. sec A= esc A. 6. sec (75° + A) = esc 2 A. 

3. tan^L = cot2 A. 7. cot (A + 50°) = tan 7 A. 

4. sin 2 A = cos 3 A. 8. sin nA = cos m A. 

5. tan^L/2 = cot2^L. 9. tan cA - cot (30° - ^1). 

10. Trigonometric ratios of 45°. Let ABC be an isosceles 
right triangle in which 

AB = 2 units, 
A = # = 45°. 
AC = BC = V2 units, 
sin 45° = V2/2 = cos 45°, 
tan 45° = 1 = cot 45°, 
fig. io sec 45° = V 2 = csc 45°. 

11. Trigonometric ratios of 30° and 
60°. Let ABD be an equilateral tri- 
angle in which AB — 2 units. Draw 
BC ±AD. 




Then 



AC = 1, 
BC = V3, 
il = 60°, 
Z ABC = 30°. 




TRIGONOMETRIC RATIOS OF ACUTE ANGLES 11 

By § 1, sin 30° = 1/2, esc 30° = 2 ; 

cos 30° = V3/2, sec 30° = 2/ V3 ; 

tan 30° = 1/ V3, cot 30° = V3. 
Also, by § 1, 

sin 60° = V3/2, esc 60° = 2/ V3 ; 

cos 60° = 1/2, sec 60° -2; 

tan 60° = V3, cot 60° = 1 /y3, or V3 /3. 

To aid the memory observe that sin 30°, sin 45°, and sin 60° are 
respectively equal to VI » V2, and V3, divided by 2. 

It is easy to read off the trigonometric ratios of 30°, 45°, and 60°, when 
we keep in mind the figures 10 and 11. 

Example. By § 9 obtain the values of the trigonometric ratios of 60° 
from those of 30°, and those of 30° from those of 60°. 

12. Approximate measurements and computations. The student 
should remember that in all actual measurements the results 
are only approximate. It is impossible to measure any 
quantity with absolute accuracy. The degree of accuracy 
sought will depend upon the importance of the results. The 
degree of accuracy secured will depend upon the instruments, 
methods, and care which are used. Likewise, in practical 
computations, a sum, difference, product, or quotient of two 
approximate values will not have a greater degree of accuracy 
than that of the least accurate of the two values. E.g., if one 
numerical measure is accurate to two figures and another to 
three figures, their sum, product, or quotient will not in general 
be accurate to more than two figures. If each of two numerical 
measures has three-figure accuracy, or if one has four-figure 
accuracy and the other only three-figure accuracy, their product 
or quotient will not in general have more than three-figure 
accuracy. The values tabulated in § 5 have only four-figure 
accuracy. 



12 PLANE TRIGONOMETRY 

13. Solving right triangles. Of the six parts (three sides 
and three angles) of a right triangle, one part (the right 
angle) is always known. If ? of any right triangle, two other 
parts are given (one at least being a side), Geometry proves 
that the triangle is entirely determined, and shows how. to 
construct it. 

Trigonometry shows how to compute the numerical values of 
the unknown parts of a triangle when the known parts are 
sufficient to determine it. 

This process is called solving the triangle. 

Hence, in solving right triangles, we must consider the two 
following cases : 

(i) Given one side and one acute angle. 
(ii) Given any two sides. 

Case (i). Ex. 1. In the right triangle ABC, A = 35° and BC = 20 feet ; 
find the numerical values of the other parts. 

Construction. To some scale (as 30 ft. to an 
inch) construct as accurately as possible a right 
q triangle having the given parts A = 35° and 
g, a= 20. 

Solution. B = 90° - A = 90° - 35° = 55°. 

Now, the ratio of either of the unknown sides 
Fig. 12 ' ' .-,.,.. x • .• £ 

to the known side a is a trigonometric ratio of 

35°, and any trigonometric ratio of 35° can be obtained from the table 

in § 5. 

Thus, b/a = cot A = cot 35°. § 1 

.-. 6/20 = cot 35° = 1.4281. by table 

.-. b = 1.4281 x 20 = 28.562. 
Again, c/20 = esc 35° = 1.7434. § 1, table 

.-. c = 1.7434 x 20 = 34.868. 

If we regard 20 as exact, or at least accurate to four figures, the values 
of b and c are accurate to only four figures ; for cot 35° and esc 35° are 
accurate to only four figures (§12). . 





SOLUTION OF TRIANGLES 13 

Check. Measure AC and AB, and multiply the number of inches in 
each by the number of feet which an inch represents. We thus obtain 
b = 28.6, c = 34.9. 

As a numerical check we could use 

a 2 = c 2 - 6 2 , or a 2 = (c + b) (c - b). 
But for simplicity and to emphasize the 
importance of accurate construction, we shall, 
in this chapter, use only the check by construc- 
tion and measurement. a - 

Fig. 13 
Case (ii). Ex. 2. In the right triangle 

ABC, BC = 83.91 ft. and AC = 100 ft. ; find the other parts. 

(A =40°, 

Given [ a ~ 83,91, to find \ B = 50°, 

^ = 10 °; [c = 130.54. 

Construct the triangle ABC having the given parts. 

rta,nA = a/b. (1) 

Formulas \ B = 90°-A. (2) 

[ c/6 = sec^4. (3) 

Computation. From (1), tan A = 83.91 / 100 

= 0.8391 = tan 40°. by table 

Hence, by § 3, A = 40°. 

From (2), B = 90° - 40° = 50°. 

From (3), c = 100- sec 40° 

= 100 x 1.3054 = 130.54. by table 

Check. By measurement A = 40°, B = 50°, AB = 131 ft. nearly. 

The solution above illustrates the five steps which, in the 
first solutions at least, should be kept separate and distinct. 

(i) Statement of the problem. 

(ii) Construction of the triangle, 
(iii) Writing the needed formulas. 
(iv) Making the computations. 

(v) Applying some check or test to answers. 



14 PLANE TRIGONOMETRY 







EXERCISE III 




Solve the right triangle ABC, 


when : 






1. A = 25°, 


a = 30. 




9. 


6 = 93.97, 


c = 100. 


2. 5 = 55°, 


6 = 10. 




10. 


a = 17.1, 


c = 50. 


3. A = 65°, 


c = 70. 




11. 


B = 75°, 


c = 40. 


4. B= 15°, 


6 = 20. 




12. 


A = 10°, 


6 = 30. 


5. £ = 35°, 


a = 50. 




13. 


A = 20°, 


c = 80. 


6. £ = 55°, 


c = 60. 




14. 


B = 25°, 


a = 30. 


7. a = 36.4, 


6 = 100. 




15. 


a = 30.21, 


c = 331. 


8. a = 23.315, 


6 = 50. 




16. 


a = 13.4, 


6=50. 



14. A vertical line at any point is the line determined by 
the plumb line at that point. 

A horizontal line (or plane) at any point is the line (or plane) 
which is perpendicular to the vertical line at that point. 

A horizontal angle is an angle whose sides are perpendicular 
to the vertical line at its vertex. 

A vertical angle is an angle whose plane contains the vertical 
line at its vertex. 

A vertical angle of which one side is horizontal is called an 
angle of elevation or an angle of depression, according as the 
second side is above or below the horizontal side. 

Note. All vertical lines converge towards the center of the earth. 
But in the next two definitions any two vertical lines are regarded as 
parallel. This is approximately true for short distances and is always 
assumed as true for such distances unless very great accuracy is required. 

The horizontal distance between two points is the distance 
from one of the two points to a vertical line through the other. 

The vertical distance between two points is the distance 
from one of the two points to the horizontal plane through 
the other. 



PROBLEMS 



15 



E.g., let MP be the vertical line at P and let the horizontal plane at 
A cut this vertical line in M ; then AM is called the horizontal distance, 
and MP is called the vertical 
distance, between the points 
A and P. Moreover Z. MAP 
is the angle of elevation of P, 
as seen from A. Also, if PN 
is horizontal at P and in the 
plane AMP, Z NPA is the 
angle of depression of A, as ^ 
seen from P. 

Now if we assume that the vertical lines at A and P are parallel, the 
lines AM and NP will be parallel also and the angles MAP and NPA 
will be equal. 




Fig. 14 



15. Solving problems. The practical problems which follow 
will illustrate the utility of the trigonometric ratios of angles 
in computing heights, distances, angles, areas, etc. 

In solving problems it will be helpful to observe the fol- 
lowing general method of procedure. 

First step. Construct accurately to some convenient scale 
a drawing which will show the relations of the given angles 
and lines to those which are required. 

Second step. Draw any auxiliary lines which may be help- 
ful in the trigonometric solution. By examining the drawing, 
fix upon the simplest steps which are necessary to solve the 
problem. 

Third step. Write the needed for- 
mulas. Make the computations, and 
check the answers. 

Ex. 1. A man, standing on the bank of a 
river at P, wishes to find how far he is from 
a tree at T on the opposite bank. He locates 
a staff at S so that PS ± PT. By measure- 
ment he finds that the horizontal distance 
PS = 250 ft., and that the horizontal angle 
PST= 40°. Find the distance PT. 




16 



PLANE TRIGONOMETRY 



By § 1, PT/PS = tan 40° = 0.8391. by table 

.-. PT = 250 ft. x 0.8391 = 209.77 ft. 
Check. By measurement PT = 210 ft. nearly, when PS = 250 ft. 

Ex. 2. A vertical flagstaff stands on a horizontal plane. At a point 
200 ft. from the foot of the staff the angle of elevation of its top is found 
to be 20°. Find the height of the flagstaff. 

Let MP (fig. 14) represent the flagstaff, and A the point from which 
the angle of elevation is taken. 

Then .<Of = 200 ft., 

and Z MAP = 20°. 

By § 1, MP /AM = tan 20° = 0.364. by table 

.-. MP = 200 ft. x 0.364 = 72.8 ft. 
Check. By measurement MP = 73 ft. nearly, when AM = 200 ft. 

Ex. 3. A man wishes to find the height of a tower DB which stands 
on a horizontal plane. From a point A on this plane he finds the angle 

£ of elevation of the top of the tower to 
be 35°. From a point C, which is in 
the horizontal plane at A and 100 ft. 
nearer the tower, he finds the angle of 
y elevation to be 65°. Find the height 
of the tower. 

Solution 1. Let DB = y ft. 
JD Then AD/y = cot 35° = 1.4281, 
and CJD/y = cot 65° = 0.4663. 

.-. 100 = AD- CD 

= (1.4281 - 0.4663) y. 
.-. y = 103.97. 

Solution 2. From C draw CE XAB, thus forming the right triangles 
ACE and CEB. 

Z CBE = Z ABD - Z CBD 

= 55° - 25° = 30°. 
CE = z ft. and CB = w ft. 
2/100 = sin 35° = 0.5736, (1) 

w/z = csc30° = 2, (2) 

and y/w = sin 65° = 0. 9063. (3) 




Now 

Let 
Then 




PROBLEMS 17 

Multiplying together (1), (2), and (3), member by member, we obtain 

2//100 = 1.1472 x 0.9063, or y = 103.97. 
Check. By measurement BB = 104 ft. nearly, when AC — 100 ft. 

Ex. 4. From the top of a hill 300 ft. higher than the foot of a tree, 
the angles of depression of the top and the foot of the tree are found to 
be 20° and 25° respectively. Find 
the height of the tree. 

Let P be the top of the hill, B 
the foot of the tree, and C its top. 
In the plane BBC draw BA hori- 
zontal at P and prolong it until it 
intersects the vertical line BC pro- 
duced in A. Fig. 17 

Then Z ABC = 20° and Z ABB = 25°. 

Let BC = x ft. 

Then AC = (300 - x) ft. 

Hence ^LP/300 = cot 25° = 2.1445, (1) 

and (300 - x) / AP = tan 20° = 0. 3640. (2) 

Multiplying together (1) and (2), member by member, we obtain 
(300 - &)/300 = 2.1445 x 0.3640. 
.\ x = 65.82 nearly. 
Check. By measurement BC = 66 ft. nearly, when BA — 300 ft. 

Note. All the problems in the following exercise need not be solved 
before beginning Chapter II. The solution of a few problems at a time, 
while the student is pursuing the more abstract and theoretic portions 
of the science, will serve to keep before him its practical utility and 
maintain his interest, 

EXERCISE IV 

1. The length of a kite string is 250 yds. and the angle of elevation 
of the kite is 40°. Find the height of the kite, supposing the line of the 
kite string to be straight. Ans. 160. 7 yds. 

2. A stick 10 ft, in length stands vertically in a horizontal area, 
and the length of its shadow is 8.391 ft. Find the angle of elevation of 
the sun. Ans. 50°. 



18 PLANE TRIGONOMETRY 

3. A tree is broken by the wind so that its two parts form with the 
ground a right-angled triangle. The upper part makes an angle of 35° 
with the ground, and the distance on the ground from the trunk to the top 
of the tree is 50 ft. Find the length of the tree. Ans. 96.06 ft. 

4. The distance between two towers on a horizontal plane is 60 ft. , and 
the angle of depression of the top of the first as seen from the top of the 
second, which is 150 ft. high, is 25°. Find the height of the first tower. 

5. At a point 200 ft. from the base of an unfinished tower, the angle 
of elevation of its top is 20° ; when completed, the angle of elevation of 
its top at this point will be 30°. How much higher is the tower to 
be built? 

6. The angle of elevation of the sun is 65° and the length of a tree's 
shadow on a level plane is 50 ft. Find the height of the tree. 

7. A chimney stands on a horizontal plane. At one point in this 
plane the angle of elevation of the top of the chimney is 30°, at another 
point 100 feet nearer the base of the chimney the angle of elevation of 
the top is 45°. Find the height of the chimney. 

8. A person standing on the bank of a river observes that the angle 
subtended by a tree on the opposite bank is 50° ; walking 40 ft. from the 
bank he finds the angle to be 30°. Find the height of the tree and 
the breadth of the river, if the two points of observation are in the same 
horizontal line at the base of the tree. 

9. The shadow of a tower standing on a horizontal plane is found to 
be 60 ft. longer when the sun's altitude is 30° than when it is 45°. Find 
the height of the tower. 

10. At a point midway between two towers on a horizontal plane the 
angles of elevations of their tops are 30° and 60° respectively. Show 
that one tower is three times as high as the other. 

11. Two observers on the same horizontal line and in the same vertical 
plane with a balloon, on opposite sides of it and 2500 ft. apart, find its 
angles of elevation to be 35° and 55° respectively. Find the height of the 
balloon. 

12. A man in a balloon observes that the bases of two towers, which 
are a mile apart on a horizontal plane, subtend an angle of 70°. If he is 
exactly above the middle point between the towers, find the height of 
the balloon, 



PROBLEMS 19 

13. From the foot of a tower the elevation of the top of a church 
spire is 55°, and from the top of the tower, which is 50 ft. high, the 
elevation is 35°. Find the height of the spire and the distance of the 
church from the tower, if both stand on the same horizontal plane. 

14. From the top of a tower whose height is 108 ft. the angles of 
depression of the top and bottom of a vertical column standing on a level 
with the base of the tower are found to be 25° and 35° respectively. 
Find the height of the column and its distance from the tower. 

15. Two pillars of equal height stand on opposite sides of a horizontal 
roadway which is 100 ft. wide. At a point in the roadway between the 
pillars the angles of elevation of their tops are 50° and 25° respectively. 
Find the height of the pillars and the position of the point of observation. 

16. A house 50 ft. high and a tower stand on the same horizontal 
plane. The angle of elevation of the top of the tower at the top of the 
house is 25°, on the ground it is 55°. Find the height of the tower and 
its distance from the house. 

17. On the top of a bluff is a tower 75 ft. high ; from a boat on the 
bay the angles of elevation of the top and base of the tower are observed 
to be 25° and 15° respectively. Find the horizontal distance of the boat 
from the tower, also the distance of the boat from the top of the tower. 

18. One of the equal sides of an isosceles triangle is 50 ft. and one of 
its equal angles is 40°. Find the base, the altitude, and the area of the 
triangle. 

19. The base of an isosceles triangle is 68.4 ft. and each of its equal 
sides is 100 ft. Find the angles, the height, and the area. 

20. The base of an isosceles triangle is 100 ft. and its height is 35.01 ft. 
Find its equal sides and the angles. 

21. The base of an isosceles triangle is 88 ft. and its vertical angle is 
70°. Find the height, the equal sides, and area. 

22. The base of an isosceles triangle is 100 ft. and the equal angles 
are each 65°. Find the equal sides, the height, and the area. 

23. The height of an isosceles triangle is 50 ft. and its vertical angle 
is 30°. Find the sides and the area. 



20 PLANE TRIGONOMETRY 

24. A man's eye is on a level with and 100 ft. distant from the foot 
of a flag pole 86.4 ft. high. When he is looking at the top of the pole, 
what angle does his line of sight make with a line from his eye to the 
foot of the pole ? 

25. A circular pond has a pole standing vertically at its center and its 
top is 100 ft. above the surface. At a point in the circumference the 
angle subtended by the pole is 20°. Find the radius and the area of the 
pond. 

26. A ladder 33-J- ft. long leans against a house and reaches to a point 
30.21 ft. from the ground. Find the angle between the ladder and the 
house and the distance the foot of the ladder is from the house. 

27. From the summit of a hill there are observed two consecutive 
milestones on a straight horizontal road running from the base of the 
hill. The angles of depression are found to be 10° and 5° respectively. 
Find the height of the hill. 

28. At the foot of a hill the angle of elevation of its summit is observed 
to be 30°; after ascending the hill 500 ft., up a slope of 20° inclination, 
the angle of elevation of its summit is found to be 40°. Find the height 
of the hill if the two points of observation and the summit are in the same 
vertical plane. 

One method of solution is similar to that of the second solution of 
example 3 in § 15. 

29. At the foot of a mountain the angle of elevation of its summit is 
35°; after ascending an opposite mountain 3000 ft., up a slope of 15° 
inclination, the angle of elevation of the summit is 15°. Find the height 
of the first mountain if the points of observation and the summit are in 
the same vertical plane. 

30. From the extremities of a ship 500 ft. long the angles which the 
direction of a buoy makes with that of the ship are 60° and 75°. Find the 
distance of the buoy from the ship, having given that cot 75° = 2 — V3- 

Ans. 125(V3 + 3)ft. 

31. There are two posts which are 240 and 80 ft. high respectively. 
From the foot of the second the elevation of the top of the first is found 
to be 60°. Find the elevation of the second from the foot of the first. 

Arts. 30°, 



PROBLEMS 21 

32. A boy standing c feet behind and opposite the middle of a football 
goal sees that the angle of elevation of the nearer crossbar is A, and the 
angle of elevation of the farther one is B. Show that the length of the 
field is c (tan A cot B — 1). 

33. A valley is crossed by a horizontal bridge whose length is I. The 
sides of the valley make angles A and B with the horizon. Show that 
the height of the bridge above the bottom of the valley is I /(cot A + cot B). 

34. Two forces of a and b lbs. respectively act in the same direction. 
Find their resultant. Illustrate the problem geometrically. 

35. Two forces of a and b lbs. respectively act in opposite directions. 
Find their resultant when a > 6, when a = 6, and when a < b. Illustrate 
each case geometrically. 

36. By two or more experiments verify that, if in any triangle ABC 
the two sides AB and BC represent two forces (both in size and direc- 
tion), the third side AC will represent their resultant, i.e. their sum in its 
simplest form. 

37. Two forces of 3 and 4 lbs. respectively act at right angles to each 
other. Show that their resultant is a force of 5 lbs. and that its line of 
action and that of the first force make an angle whose tangent is 4/3. 

Suggestion. In fig. 4 let OM and MP respectively represent the two 
forces, then the line OP will represent the resultant. 

38. Two forces of a and b lbs. respecti vely act at right angles. Show 
that their resultant is a force of Va 2 -f b 2 lbs. , and that its line of action 
and that of the first force make an angle whose tangent is b/a. 

39. Two forces act at right angles. The first is a force of 3 lbs. and 
the resultant is one of 5 lbs. Show that the second force is one of 4 lbs., 
and that the lines of action of the first force and the resultant form an 
angle whose cosine is 3/5. 

40. Two forces act at right angles. The first is a force of a l bs, and the 
resultant is one of c lbs. Show that the second force is one of Vc 2 — a 2 lbs. 
and that the lines of action of the first force and the resultant form an 
angle whose cosine is a/c. 



CHAPTEE II 



TRIGONOMETRIC RATIOS OF POSITIVE AND NEGATIVE 
ANGLES OF ANY SIZE 

16. Positive and negative angles of any size. In the first 
chapter we studied acute angles and considered their size only. 
When, however, we conceive an angle as generated by a rotat- 
ing line, we see that it can be either positive or negative and 
of any size whatever. 

Thus, suppose a line OP to start from OX and to rotate 

about counter-clockwise; 
that is, in a direction oppo- 
site to that of the hands of 
a clock. 

When OP reaches the 
position OP x it has gener- 
ated the acute angle XOP x . 
When OP reaches the posi- 
tion OP 2 it has generated the 
obtuse angle XOP 2 . When 
OP reaches OP 3 it has gener- 
ated the angle XOP s (i.e. Z XOP 2 + Z P 2 OP 3 ). When OP 
reaches OP 4 it has generated the angle XOP± (i.e. Z XOP 3 
+ Z P3OP4). When OP reaches OX it has generated an angle 
of 360°. 

If OP continues to rotate, when it reaches OP x the second 
time it has generated the angle 360° + the acute angle XOP x ; 
when OP reaches OP x the third time it has generated the angle 
720° + the acute angle XOP x ; and so on for any number of 
revolutions. 

22 




TRIGONOMETRIC RATIOS 23 

When the rotation of OP is counter-clockwise, the angle 
generated is said to be positive ; hence, when the rotation of 
OP is clockwise, the angle generated is negative. 

E.g., in ten minutes the minute hand of a clock generates a negative 
angle of 60° ; 

in 15 minutes it generates an angle of — 90° ; 
" 30 " " " " " " - 180°; 

" 1 hour " " " " " - 360°; 

" 3J hours " " " " " -(3 x 360°+ 180°); 

and so on. If the hands of a clock were to rotate in the opposite direction, 
i.e. counter-clockwise, they would generate positive angles. 

The line OX which marks the first position of the rotat- 
ing line OP is called the initial side of the angle XOP z ; and 
the line OP 3 which marks the final position of OP is called the 
terminal side of this angle. 

The size of an angle gives the amount which its generating 
line has rotated, and its quality* gives the direction of this 
rotation. 

The value of a positive or a negative angle includes both its 
size and its quality as positive or negative. 

17. Coterminal angles. Any angle, positive or negative, 
which has the same initial side and the same terminal side 
as angle A is said to be coterminal with A. If any angle, as 
XOP 2 in fig. 18, is increased or diminished by 360° (or by any 
entire multiple of 360°), the resulting angle, whether positive 
or negative, will have the same initial and the same terminal 
side as XOP 2 . 

* In Algebra the quality of a particular number as positive or negative is 
denoted by the sign + or — , and this quality is often called the sign of the 
number. It is unfortunate, however, to use the same word sign as the name 
both of a symbol and also of the property of number denoted by this symbol. 
Moreover the introduction of the word sine adds another reason for not calling 
the quality of a number its sign in Trigonometry. 



24 PLANE TRIGONOMETRY 

Hence if n is any integer, positive or negative, then all the 
angles, and only those, which are or can be made coterminal 
with any angle A are denoted by n 360° + A. E.g., 2 • 360° + 40° 
is or can be made coterminal with 40°. 

Evidently there are as many different angles coterminal 
with A as there are different entire values for n. 

18. Quadrants. If, as in fig. 18, the initial side of the angle 
XOP 1 is produced through the vertex to X\ and the perpen- 
dicular YOY' drawn through the vertex 0, these lines will 
divide the plane of the angle into four equal parts called 
quadrants. These quadrants are numbered in the positive 
direction, reckoning from the initial side of the angle under 
consideration ; that is, if OX is the initial side of the angle 
considered, then XOY will be the first quadrant; YOX 1 the 
second quadrant; X'OY 1 the third quadrant; and Y'OX the 
fourth quadrant. 

If OF is the initial side of the angle considered, then YOX f 
will be the first quadrant ; and so on. 

For convenience, an angle is said to be in (or of) that 
quadrant in which its terminal side lies. 

E.g., the angle XOP 2 (fig. 18) is said to be in the second quadrant, 
since its terminal side OP 2 lies in that quadrant ; the angle XOP± is said 
to be in the fourth quadrant, since its terminal side OP± is in that quad- 
rant. The angle YOP 2 is in the first quadrant, and YOP3 is in the 
second quadrant, since here OY is the initial side and YOX' is the first 
quadrant. 

Again, 200° = 180° + 20°, hence an angle of 200° is in the third quad- 
rant ; 880° = 2 (360°) + 160°, hence an angle of 880° is in the second 
quadrant. An angle of — 50° is in the fourth quadrant, and an angle of 
- 330° is in the first quadrant. Since - 400° = - 360° - 40°, an angle 
of — 400° is in the fourth quadrant. 

19. Two angles are said to be complementary when their 
sum is 90° (§ 8), and supplementary when their sum is 180°. 



DIRECTED LINES 



25 



E.g., the complement of 110° is 90° - 110°, or - 20°; 

" » «_80°" 90° -(-80°), or 170°; 

u u u ^ u 90° -J.; 

and u " " -A" 90°- (- A), or 90° + A. 

" supplement " 135° " 180° - 135°, or 45°; 

44 " " 235° " 180° - 235°, or - 55°; 

u « u ^ « 180°- A; 

and " " " -A u 1$0 o -{-A),ot1S0° + A. 



EXERCISE V 



In which quadrant is each of the following angles ? 



7. -225°? -300°? 

8. -415°? -842°? 

9. 942° ? - 1174° ? 



1. 5/3 right angles? 4. 150°? 317°? 

2. 3f right angles ? 5. 847°? 1111°? 

3. 17£ right angles ? 6. - 35° ? - 140° ? 

10. Construct the angles in examples 5, 7, 9. 

11. Give two positive and two negative angles, each of which is coter- 
minal with 45° ; 30° ; 100° ; 200° ; - 10° ; - 100°. 

Find the complement and the supplement of : 

12. 165°. 14. 295° 17' 14". 16. - 32° 14' 21". 

13. 228°. 15. 314° 22' 17". 17. - 165° 28' 42". 
Find the smallest positive angle co terminal with : 

18. 420°. 19. 895°. 20. -330°. 21. -740°. 22. -1123°. 

20. Positive and negative lines. If two lines extend in 
opposite directions and one of them is regarded as positive, 
the other will be negative. A positive or a negative line is 
called a directed line, and is read in the direction in which it 
extends or is supposed to be traced. 

< C 



D- 



FiG. 19 

Of the directed line AB, A is called the origin and B the end. 



26 



PLANE TRIGONOMETRY 



E.g., as a directed line, AB extends from its origin A towards its enc 
B, and CD extends from its origin C towards its end D. 

If we call AB positive, BA or CD will be negative. 

Hence AB=— BA, or AB + BA = 0. 

The numerical measure of a positive or a negative line is a 
positive or a negative real number. E.g., if AB is four units 
in length and is regarded as positive in direction, then 
AB = + 4 units and i?^l = — 4 units. 

21. Trigonometric ratios of positive or negative angles of any 
size. In each of the four figures below, let A denote any 
angle, positive or negative, which is coterminal with the angle 
XOP. In each figure a curved arrow indicates the smallest 
positive value of A, and a dotted arrow the smallest in size of 
its negative values. 




>X 



>x 



Fig. 20 

From any point in the terminal side OP, as P, draw MP 
perpendicular to the initial side OX or OX produced through 0. 

In each of the four figures we have three directed lines, OM, 
OP, and MP. The origin of the directed line OM or OP is at 
the vertex of the angle, and the origin of MP is in the initial 
side of the angle or in that side produced. 



TRIGONOMETRIC RATIOS 27 

OM is regarded as positive when it extends in the direction 
of the initial side of the angle, OX ; and hence it is negative 
when it extends in the opposite direction, 0X f . Thus OM is 
positive in fig. a or d, and negative in fig. b or c. 

MP is regarded as positive when it extends upward, or into 
the first or second quadrant ; hence it is negative when it 
extends downward, or into the third or fourth quadrant. Thus 
MP is positive in fig. a or b, and negative in fig. c or d. 

OP in every position extends in the direction of the terminal 
side of the angle and is regarded as positive. 

Observe that in each figure P is a point in the terminal 
side ; MP gives the distance and direction of P from the initial 
side OX, and OM gives the distance and direction of MP from 
the vertex 0. 

Ex. 1. What is the quality of MP and OM respectively when A is 
in the first quadrant ? the second quadrant ? the third quadrant ? the 
fourth quadrant ? 

Ex. 2. The angle A is in one of which tw T o quadrants when MP is 
positive ? MP negative ? OM positive ? OM negative ? 

The six simple ratios (three ratios and their reciprocals) 
which can be formed with the three directed lines, MP, OM, 
and OP, are called the trigonometric ratios of the angle A. 

The following definitions do not differ from those in § 1 
except in their generality, which follows from the use of 
positive and negative angles and lines. 

The ratio MP / OP is the sine of A ; 

and its reciprocal OP / MP is the cosecant of A. 

The ratio OM / OP is the cosine of A ; 

and its reciprocal OP / OM is the secant of A. 

The ratio MP/OM is the tangent of A ; 

and its reciprocal OM/MP is the cotangent of A. 



28 PLANE TRIGONOMETRY 

If two angles are or can be made coterminal, any trigo- 
nometric ratio of the one is evidently equal to the same 
trigonometric ratio of the other. 

Since any angle denoted by n-360° + A, where n is any 
real integer, can be made coterminal with A, it follows that 

Any trigonometric ratio of (n • 360° + A) is equal to the same 
trigonometric ratio of A, 

Example. Eind a positive acute angle whose trigonometric ratios are 
equal to those of 420° ; 760°; 1120°; -340°; -660°. 

Since 1120°, or 3 ■ 360° + 40°, is coterminal with 40°, any trigono- 
metric ratio of 1120° is equal to the same ratio of 40°. 

22. Laws of quality of the trigonometric ratios. Two recip- 
rocal trigonometric ratios must evidently have the same 
quality. 

Since OP is always positive, the reciprocal ratios sin A and 
esc A have the same quality as MP. 

Hence sin A or esc A is positive when A is in the first or the 
, second quadrant, and negative when A is in the third or the 
fourth quadrant. 

The reciprocal ratios cos A and sec A have the same quality 
as OM. 

Hence cos A or sec A is positive when A is in the first or the 
fourth quadrant, and negative when A is in the second or the 
third quadrant. 

The reciprocal ratios tan A and cot A are positive or nega- 
tive according as MP and OM are like or opposite in quality. 

Hence tan A or cot A is positive when A is in the first or 
the third quadrant, and negative when A is in the second or the 
fourth quadrant. 

Observe that when A is in the first quadrant all its trigo- 
nometric ratios are positive, and when A is in any other 
quadrant only tivo of its six ratios are positive, and these two 
are reciprocals. 



TRIGONOMETRIC RATIOS 29 

The figure below, where XOY is the first quadrant, may 

help to fix in mind the very important laws of quality given 

above. 

Y 



sin and esc + 



tan and cot + 



All the ratios -f 



cos and sec -f- 



Fig. 21 

E.g., the angle 500° is in the second quadrant ; hence all its trigono- 
metric ratios are negative except its sine and cosecant. The angle — 300 c 
is in the first quadrant ; hence all its ratios are positive. 

Example. What is the quality of each trigonometric ratio of 103°? 
-135°? 235°? -75°? 325°? -325°? 660°? 1100°? 

23. Sin _1 c, cos -1 c, .... If sin A = c, then 
A = any angle whose sine is c. 

The customary expression for any angle whose sine is c is 
sin -1 c, read any angle whose sine is c, or briefly, angle sine c. 

Thus, if sin A = c, A = sin _1 c, and conversely. 

A similar meaning is given to the expressions, cos _1 6, tan _1 «, 
cot -1 a, sec _1 A, csc -1 &. 

E.g., sin~ 1 (l/2) denotes any angle whose sine is 1/2 ; hence it denotes 
any angle which is coterminal with 30° or 150° ; that is, 

sin-i(l /2) = n • 360° + 30° or n • 360° + 150°, 

where n is any integer, positive or negative, including 0. § 17 

Again, tan" 1 1 = n • 360° + 45° or n ■ 360° + 225°. 

Ex. 1. If A = cotr- 1 (— 1), what are the values of A ? 

Ex. 2. If A = cos-^- 1/2), what are the values of A ? 

Ex. 3. Given A = sin- 1 (4/5), to construct A and find its other trigo- 
nometric ratios. 

Since sin A is -f , the angle A is in the first or the second quadrant. 



30 



PLANE TRIGONOMETRY 



Draw OX, at draw OY J_ OX, and lay off OB equal to 4 units. 
Through B draw P'P parallel to OX. From O as a center and with 

a radius equal to 5 units describe an 
arc cutting P'BP in some points, as 
P' and P. Draw OP and OP'. Then 
^1, or sin- 1 (4/5), is any angle which is 
coterminal with XOP or XOP'. 

Hence A, or sin— x (4/5), is the acute 
angle XOP + n • 360°, or the obtuse 
angle XOP' -f n • 360°, where n is any 
integer, positive or negative, including 
zero. 




Here 
Hence 



and 



OM = + 3 and OW = - 3. 

sin -4 = 4/5, cscJ. = 5/4; 

cos J. = ±3/5, secJ. = ± 5/3 ; 

tan-4 =± 4/3, cot A =± 3/4. 



§21 



When, as above, any trigonometric ratio of A has two values which 
are written together, we shall consider the upper sign as belonging to the 
trigonometric ratio of the least positive value of A. Thus, if A is in the 
first or the second quadrant, we shall write tan A = ± 4/3 ; while if A 
is in the second or the third quadrant, we shall write tan A = =f 4/3 ; 
and so on. 



EXERCISE VI 

Construct A and find its other five trigonometric ratios when : 

1. ^. = sin-i(-2/3). 7. A = cos-^- 3/7). 

2. 4 = tan- 1 (5/2). 8. A = cot- 1 (6/3). 

3. A = t&n-i(-S). 9. A = cos- 1 (-4/5). 



4. A = cos- 1 (2/3). 



10. A msec- 1 2. 



5. A = sin- 1 (-7/8). 11. A = sec- 1 *- 3/2). 

6. A= tan- 1 7. 12. ^4 = csc-^- 5/3). 

13. Express each of the trigonometric ratios of A in terms of sin A. 
If sin A is positive, A is in the first or the second quadrant. 
In fig. 22, let OP = 1. 

Then sin A is the measure of MP or 1TP 7 . 



FUNDAMENTAL RELATIONS 31 

Whence MP = M'P' - sin A, 



0M = Vl -sin 2 ^4, OM' = - Vl - sin 2 .4. 



Hence cos -4 = ± Vl — sin 2 ^4, sec .4 = ± 1 /Vl — sin 2 ^. ; 

tan A = ± sin4./Vl — sin 2 .4, cot A — ± Vl — sin 2 4./sin.4. 
If sin .4 is negative, A is in the third or the fourth quadrant, and 

cos 4. ==f Vl - sin 2 A, sec A - =f 1 / Vl - sin 2 A. 

24. Fundamental relations between the trigonometric ratios of 
any angle A. 

From the definitions of the trigonometric ratios of A , we have 

sin A esc A = 1, cos A sec A = 1, tan A cot A = 1. [1] 
MP MP I OP sin A 

tan a — ~e=. - - = • r^n 

OM OM/OP cos A L ~ J 

Taking the reciprocals of the members of T2]. we obtain 

cot A = cos A / sin A. [3] 

In each of the figures in § 21, we have 

MP' 2 + ~OM* -Tip 2 . (1) 

Dividing the numbers of (1) by OP , we obtain 

(MP /OP) 2 + (OM/ OP) 2 = 1. (2) 

If, for brevity, we write (sin .I) 2 and (cos .I) 2 in the form 
sin 2 .! and cos 2 . 4, from (2) we obtain 

sin 2 A + cos 2 A = 1. [4] 

Dividing the members of (1) by OM", we obtain 
(MP /OM) 2 + 1 = (OP /OM) 2 , 
or tan 2 A + 1 = sec 2 A. [5] 

Dividing the members of (1) by MP , we obtain 
1 ^(OM /MP) 2 = (OP /MP) 2 , 
or cot 2 A + 1 = esc 2 A. [6] 



32 PLANE TRIGONOMETRY 

The identities [1] • • • [6] express the more important of the 
numberless relations that exist between the trigonometric ratios 
of any angle A. 

For brevity (sinJ.) w , (cos^4) n , etc., are written in the form 
sin w ^4 ? cos n .4, etc., as above, except when n = — 1. 

By § 23, sin -1 c is used to denote any angle whose sine is c; 
hence the reciprocal of sin A should never be written in the 
form sin _1 ^4, but in the form (sin ^4) _1 or 1/sinA 

Ex. 1. State identities [1] • • • [6] in words. 

Ex. 2. sec A = — 4 ; find the values of the other ratios of A. 

Since sec A is — , A is in the second or the third quadrant. § 22 

sec A = — 4 ; .-. cos A = — 1 /4. by [1] 

sin A = ± Vl - cos 2 A by [4] 
= ± Vl -1/16 = ± V15/4. 

.-. esc A = ± 4/V15 = ± 4 V15/15. by [1] 

tan A = sin A /cos A =T V15. by [2] 

cot^L=Tl/V15= =F V15/15. by [1] 

Check. Construct ^4 from sec A = — 4, and then find the other trigo- 
nometric ratios of A, as in § 23. 

Ex. 3. Express the other trigonometric ratios of A in terms of sin A. 

esc A = 1 /sin A. by [1] 

cos A = ± Vl -sin 2 J.. by [4] 

.-. sec A = ± 1 / Vl -sin 2 .4. by [1] 

tan A = sin A /cos A by [2] 
= ± sin A / Vl — sin 2 A. 

.-. cot J. = ± Vl - sin 2 A /sin ^4. by [1] 

When sin A is positive A is in the first or the second quadrant ; when 
A is in the first quadrant all the trigonometric ratios of A are + ; when 
A is in the second quadrant only sin A and esc A are -f . The signs as 
written above are for sin A positive. 

Check. Find these relations as in example 13 of Exercise VI. 



PROOFS OF IDENTITIES 33 

EXERCISE VH 

By § 24, compute the other trigonometric ratios of A, having given : 

1. sin -4. =-2/3. 5. tan A =-4/3. 9. esc A = - V3. 

2. cosA = l/S. 6. cot A =-2. 10. sec A = 4. 

3. sin ^4=0.2. 7. cotA = S/2. 11. tan^=-V7. 

4. cos^4=-3/4. 8. tan -4 = 2.5. 12. cos A = m/c. 

Express each of the trigonometric ratios of A in terms of : 

13. cos J.. 14. tanX 15. cot A. 16. sec A. 17. csc^4. 

25. Proofs of identities. Of the different ways of proving 
an identity, the three following are the more common and 
important. 

(i) Derive the required identity from one or more known 
identities. 

Ex. 1. Prove that ± Vsec 2 A + esc 2 A = tan A + cot A. (1) 

Adding identities [5] and [6] in § 24, we obtain 

sec 2 A + esc 2 A = tan 2 A + 2 + cot 2 A 

= (tan A + cot ^4) 2 . by Algebra, [1] 

Extracting the square root of both members of the last identity, we 
obtain identity (1). 

(ii) Reduce one member of the required identity to the form 
of the other member, using any known identities. 



; Vrr^ 



■ cos A. 
Ex. 2. Prove that a I- = esc A — cot A. 

■ cos A 



esc A - cot A = — ^4 »y PL ffl 

sin ^4 sin A 

1 — cos ^4 

Vl — cos 2 A 



by Algebra, [4] 

V (l - cos AY _ /l - cos ^4 
1 -cos 2 -4 ~" \l + cos4' y ge ^ 



34 



PLANE TRIGONOMETRY 



(iii) Reduce one member to its simplest form,, and then reduce 
the other member to the same form. 

When an identity contains any other trigonometric ratios 
than the sine and cosine, it is usually best in this method to 
replace these other ratios by their values in terms of the sine 
and cosine. 



Ex. 3. Prove that 

sin 2 A tan A + cos 2 A cot A + 2 sin A cos A ; 



i tan A -f cot A. (1) 



^. , . „ a sinJL . cos A _ . . 

First member = sin 2 A h cos 2 A f- 2 sin A cos A 

cos A sin A 

_ sin 4 A + cos* A + 2 sin 2 ^L cos 2 J. 

sin J. cos ^i 
_ (sin 2 A + cos 2 A) 2 

sin J. cos ^1 
= l/(sin A cos ^4). 
Similarly show by [3], [4], and Algebra, that 

second member = 1 / (sin A cos A). 
From the last two identities we obtain identity (1). 



by Algebra 

by Algebra 

by [4] 



EXERCISE VIII 

Prove each of the following identities : 

1. cos A tan A = sin A. 

2. sin A sec A = tan A. 

3. cos A esc A = cot A. 

4. sin A cot A = cos A. 

5. cos 2 .4 — sin 2 A = 1 — 2 sin 2 J.. 

6. cos 2 A - sin 2 J. = 2 cos 2 J. - 1. 

sin A _ 1 — cos A 
1 + cos A sin A 

1 + sin A _ cos A 



Obtain from [2] 
From [2] by [1] 
From [3] by [1] 

From [4] by Algebra 
From [4] by Algebra 



8. 



cos^. 



1 — sin A 



IDENTITIES 



35 



9. 



sec A -f 1 _ tan A 



From [5] by Algebra 



From [5] by [1] 
From [2] by [1], [5] 



tan A ~ sec A — 1 

10. sec A + tan A = 1 / (sec A - tan ^4). 

11. (1 + tan 2 ^4) cos 2 ^4 = 1. 

12. (1 + cot 2 ^4)sin 2 ^4 = l. 

13. sin 2 A + sin 2 A tan 2 A = tan 2 A . 

14. (csc 2 ^4 — l)csc 2 ^4 = cos 2 J.. 

15. cos 4 ^4 - sin 4 A + 1=2 cos 2 ^4. 

cos 4 A - sin 4 A +1 = (cos 2 A + sin 2 A) (cos 2 A - sin 2 A) + 1 
= cos 2 ^L + (1 - sin 2 ^4) = 2 cos 2 .A. 

16. tan 2 A/(l + tan 2 4 ) = sin 2 A . 

cos A 



17. 



Vl - sin 2 A 



sinA Vl - cos 2 ^4 

18. cot 2 ^4 - cos 2 A = cot 2 A cos 2 ^4 . 

19. sec 2 ^4 + csc 2 ^L = sec 2 A esc 2 ^4. 

20. tan A + cot A = sec ^4 esc A. 
cot J. cos J. cot A - cos .4 



21 



cot A + cos J. cot .4. cos J. 

sec 2 A + esc 2 J. 



22. tan A + cot ^4 ; 



sec A esc J. 



23. l/Vsec 2 ^L - 1 = Vcsc 2 J. - 1. 
1 + tan 2 A _ sin 2 ^4 esc A 



24. 
25. 



1 + cot 2 A cos 2 A ' cot A -f tan J. 
cos A sin ^4 



ecos^.. 



= sin A -f cos ^4. 



1 — tan ^4 1 — cot J. 

26. sin 3 A cos ^4 + cos 3 ^4 sin A = sin A cos J.. 

27. sin 2 A cos 2 A + cos 4 A = 1 - sin 2 ^4. 
sin ^4 



28. J 1 " 8111 - 1 
\ 1 + sin ^. 



=.sec A — tan ^4. 



36 



PLANE TRIGONOMETRY 



29. sinA + 1 + COsA =2cscA. 
1 + cos A sin A 

30. 1 / (cot A -f tan A) = sin A cos A. 

31. 1 / (sec A — tan A) = sec A -f- tan A. 
1 — tan A cot ^4 — 1 



32. 



1 + tan A cot A -f- 1 



33. ^^^ EECos^-sin^l. 
1 + tan 2 .A 

34. esc A / (cot A + tan A) = cos A 

35. csc 4 J.(l - cos 4 J.) -2cot 2 A = l. 

26. Changes of the trigonometric ratios of A as A increases from 
0° to 360°. To simplify this discussion, let OP have the same 
length in all of its positions. 

Changes of sin A. Let A = XOP, and let OP revolve coun- 
ter-clockwise about from the position OX. 



x'4f 




CHANGES OF TRIGONOMETRIC RATIOS 37 

While A increases from 0° to 90°, 

MP increases from to OP ; 
hence MP / OP, or sin A, increases from to + 1. 
While A increases from 90° to 180°, 

MP decreases from OP to ; 
hence MP / 'OP, or sin A, decreases from + 1 to 0. 
While A increases from 180° to 270°, 

MP decreases from to — OP ; 
hence MP j OP, or sin A, decreases from to — 1. 
While ' A increases from 270° to 360°, 

MP increases from — OP to ; 
hence MP j OP, or sin A, increases from — 1 to 0. 

Changes of cos A . . 

While A increases from 0° to 90°, 

OM decreases from OP to ; 

hence OM / OP, or cos A, decreases from + 1 to 0. 

While A increases from 90° to 180°, 

OM decreases from to — OP ; 

hence OM J OP, or cos A, decreases from to — 1. 

Similarly the pupil should obtain the other changes of cos A 
given in the table below. 

Changes of tan A. Let XOP approach 90° or 270° (either 
from a less or a greater value) so that OM decreases in 
size to 1/2 its value the first second of time, to 1/2 its 
remaining value the next second, to 1/2 its second remaining 
value the third second, and so on indefinitely. Then, since 
MP increases slightly, MP / OM, or tan A, more than doubles 
its value the first second of time, more than doubles its new 
value the next second, more than doubles its last value the 
third second, and so on indefinitely. Thus tan A will exceed 
in arithmetic (or absolute) value any assignable constant num- 
ber however great ; that is, when A approaches very near 90° 
or 270°, tan A = + oo or — oo. 



38 



PLANE TRIGONOMETRY 



Also, when A = 0°, 180°, or 360°, MP = 0, and therefore 
tan A = 0. 

Hence we have the changes of tan A found in the table below. 

Changes of cot A. Let X OP approach 0°, 180°, or 360° 
(either from a* less or greater value) so that MP decreases 
in size to 1/2 its value the first second of time, to 1/2 its 
new value the next second, and so on ; then OM/MP, or 
cot A, becomes + oo or — oo ; that is, when A approaches near 
0°, 180°, or 360°, cot A = + oo or - oo. 

Also, when A = 90° or 270°, OM = 0, and therefore cot A = 0. 

Hence we have the changes of cot A given in the table below. 

Similarly the changes of sec A and esc A, which are tabu- 
lated below, should be proved by the student. 

By remembering that two reciprocal numbers are like in 
quality, that when the one increases the other decreases, and 
that their corresponding values are reciprocals of each other, 
the changes of esc A, sec A, and cot A are known from the 
changes of sin A, cos A, and tan A respectively. 



A increases from 


0° to 90° 


90° to 180° 


180° to 270° 


270° to 360° 


sin A varies from 


to +1 


+ 1 to 


to - 1 


- 1 toO 


esc A ' ' " 


+ 00 to +1 


+ 1 tO + OO 


— oo to — 1 


— 1 to — 00 


cos A " " 


+ 1 toO 


to - 1 


- 1 to 


to + 1 


sec .4 " " 


+ 1 to +OD 


— oo to — 1 


— 1 tO — 00 


+ oo to + 1 


tan A increases from 


to +oo 


— oo to 


to + 00 


— oo to 


cot A decreases " 


+ oo to 


to — oo 


+ oo to 


to — oo 



From what precedes it follows that : 

The tangent or the cotangent can have any real value. 

The sine or the cosine can have any value from — 1 to + 1 
inclusive. 

The secant or the cosecant can have any value from — oo to 
— 1 or from + 1 to + oo inclusive. 



RATIOS OF 0°, 90°, 180°, 270° 



39 



Observe that neither the sine nor the cosine can have a value greater 
than + 1 or less than — 1 ; and that neither the secant nor the cosecant 
can have any value between — 1 and -f 1. 

E.g., -f 3/4 is the sine of some angle, the cosine of some angle, the 
tangent of some angle, or the cotangent of some angle ; but it can be 
neither the secant nor the cosecant of any angle. Again, — 3/2 can be 
neither the sine nor the cosine of any angle. 

27. Trigonometric ratios of 0°, 90°, 180°, 270°. When A = 90° 
or 270°, MP = + OP or - OP and OM = ; hence tan .4 or 
sec A assumes the form ± OP/0. Now the division of OP by 
zero is impossible ; hence, strictly speaking, 90° or 270° has no 
tangent or secant. But when A approaches very near to 90° 
or 270°, by § 26 tan A or sec A is + oo or — go ; hence it is 
customary to say that the tangent or secant of 90° or 270° is oo, 
meaning thereby that however near A approaches to 90° or 
270°, tan A or sec A is + cc or — oo. 

Again, when A = 0° or 180°, MP = and OM = + OP or 
— OP] hence cot .4 or esc J assumes the form ± OP/0. 
Therefore, strictly speaking, 0° or 180° has no cotangent or 
cosecant. But when A approaches very near to 0° or 180°, 
by § 26 cot A or esc A is + oo or — oo ; hence it is customary 
to say that the cotangent or cosecant of 0° or 180° is oo. 

The trigonometric ratios of 0°, 90°, 180°, and 270° are tabu- 
lated below. To aid the memory, the reciprocal ratios are 
grouped together. 



Angle 


0° 


90° 


ISO 


270° 


sine 





+ 1 





- 1 


cosecant 


00 


+ 1 


CO 


- 1 


cosine 


+ 1 





- 1 





secant 


+ 1 


CO 


- 1 


00 


tangent 





GO 





00 


cotangent 


GO 





00 






40 



PLANE TRIGONOMETRY 



Note. Putting OP = a, tan 90° assumes the form a/0, where a ^ 0. 
The form a/0 could be used as the tangent of 90°; then, whether we 
regarded a/0 as a symbol without numerical meaning, as a symbol of 
impossibility, or as a symbol of absolute infinity, when a/0 appeared as 
the tangent of A, the value of A would be known as definitely as when 
the tangent of A is any finite number. 

28. The trigonometric ratios of — A in terms of the ratios of A. 




>X 



Fig. 24 

In each figure let A denote any angle, positive or negative, 
which is coterminal with XOP ; then — A will be coterminal 
with XOP'. Angle A is in the first quadrant in fig. a, in the 
second quadrant in fig. b ; and so on. 

Take OP = OP', and draw PM ± OX and P'M' _L OX. 

Then in each figure the acute angles MOP and M'OP' will 
be equal in size. Hence any two corresponding sides of the 
A OMP and OM'P' will be equal in length. Therefore, as 
directed lines, 

M'P' J OP' = - MP J OP, i.e. sin (- A) = - sin A, (1) 
and OM' I OP' = OM/ OP, i.e. cos (— A) = cos A. (2) 



TRIGONOMETRIC RATIOS OF - A 41 

Dividing (1) by (2), tan(- A)=-faiiA. 
Dividing (2) by (1), cot (- A) = - cot A. 

From (2) by [1], sec (— A) = sec ^4. 
From (1) by [1], esc (— A) = — esc A. 

Identity (1) states that sin(— A) and sin A are arithmetic- 
ally equal but opposite in quality ; that is, when sin(— ^1) 
is — , sin A is + ; and when sin (— .4) is +, sin A is — . 

Identity (2) states that cos (—A) and cos A are arithmetically 
equal and like in quality. 

The six identities just proved can be summed up as follows : 

Any trigonometric ratio of — A is equal arithmetically to the 
same ratio of A ; hut only the cosines (or the secants) of — A 
and A are like in quality. 

E. g. , sin ( - 35°) = - sin 35°, cos ( - 98°) = cos 98°, 

tan (-212°) = - tan 212°, esc (-317°) = - esc 317°. 

Ex. 1. Express each trigonometric ratio of — 22° in terms of a ratio 
of 22°. 

Ex. 2. Express each trigonometric ratio of 320° in terms of a ratio of 
a positive angle less than 45°. 

An angle of 320° is coterminal with one of — 40° ; hence any trigono- 
metric ratio of 320° is eqnal to the same ratio of — 40° (§ 21). 

Whence sin 320° = sin (- 40°) = - sin 40°, 

cos 320° = cos (- 40°) = cos 40°, 
tan 320° - tan (- 40°) = - tan 40°, etc. 

Similarly the trigonometric ratios of any angle in the fourth quadrant 
can be fonnd in terms of those of some positive acute angle. 

Ex. 3. Express each trigonometric ratio of — 325° in terms of a ratio 
of a positive angle less than 45°. 

An angle of — 325° is coterminal with one of 35° ; hence 
sin (- 325°) = sin 35°, cos (- 325°) = cos 35°, etc. 

Similarly the trigonometric ratios of any angle in the first quadrant 
can be found in terms of those of some positive acute angle. 

29. The trigonometric ratios of 90° + A in terms of the ratios 
of A. In each figure let A denote any angle, positive or 



42 



PLANE TRIGONOMETRY 



negative, which is coterminal with XOP, and let POP' = 90° 
then A + 90°, or 90° + A, is coterminal with XOP 1 . 




Fig. 25 



Take OP = OP', and draw PM _L OX and P'il/' _L OX. Then 
the acute angles MOP and M'P'O will be equal in size. Hence 
any two corresponding sides of the A MOP and M'OP 1 will be 
equal in length. Therefore, as directed lines, 

M'P'/OP' = OMj OP, i.e. sin (90° + A) = cos A ; (1) 

and OM'/OP' = - MP /OP, i.e. cos (90° + A) = - sin A. (2) 

.*. tan (90° + .4) = - cot ,4, cot (90° + A) = — tan ,4, 

sec (90° + A) = - esc A, esc (90° + A) = sec A. 

Since the angle A is 90° less than the angle 90° + A, the 
six identities just proved can be summed up as follows : 

Any trigonometric ratio of an angle is equal arithmetically 
to the co-ratio of this angle less 90°, but only the sine {or the 
cosecant) of the first angle has the same quality as the co-ratio 
of the second angle. 



TRIGONOMETRIC RATIOS OF 90° + A 48 

E.g., since 130° - 90° = 40°,' we have 

sin 130° = cos 40°, tan 130° = - cot 40° ; 
cos 130° = - sin 40°, cot 130° = - tan 40°. 

Ex. 1. Express in terms of a trigonometric ratio of some positive 
angle less than 45° each trigonometric ratio of 126° ; 492° ; — 220°. 

sin 126° = cos 36°, cos 126° = - sin 36°, etc. § 29 

An angle of 492° is coterminal with one of 132° ; hence 

sin 492° = sin 132° = cos 42° ; §§ 21, 29 

cos 492° = cos 132° = - sin 42° ; etc. 
An angle of — 220° is coterminal with one of 140° ; hence 

sin (- 220°) = sin 140° = cos 50° = sin 40 ; §§ 21, 29, 9 

cos (- 220°) = cos 140° = - sin 50° = - cos 40° ; etc. 

Similarly the trigonometric ratios of any angle in the second quadrant 
can be found in terms of those of some positive acute angle less than 45°. 

Ex. 2. Express in terms of a trigonometric ratio of some positive 
angle less than 45° each trigonometric ratio of — 130° ; 230°. 

sin (- 130°) = - sin 130° = - cos 40° ; §§ 28, 29 

cos (- 130°) = cos 130° = - sin 40° ; 
tan (- 130°) = - tan 130° = cot 40°. 
Applying § 29 twice in succession and then § 9 once, we have 
sin 230° = cos 140° = - sin 50° = - cos 40° ; 
cos 230° = - sin 140° = - cos 50° = - sin 40° ; 
tan 230° = - cot 140° = tan 50° = cot 40° ; etc. 
Similarly we can find, in terms of the trigonometric ratios of a positive 
angle less than 45°, the ratios of any positive or negative angle in the 
third quadrant. 

The principles in §§9, 28, 29 have an important bearing 
on the construction and nse of trigonometric tables and on 
the solution of triangles. By them, as is seen above, the 
trigonometric ratios of any angle can be expressed in terms of 
the trigonometric ratios of some positive angle less than 45°. 
Hence, from a table which contains the trigonometric ratios of 
all angles between 0° and 45°, we can obtain the trigonometric 
ratios of any angle whatever. 



44 PLANE TRIGONOMETRY 

30. Trigonometric ratios of complementary and supplementary 
angles. Applying § 29 twice and then § 28 once, we obtain 

sin (180° -A)= cos (90° - .4) = - sin (- A) = sin A ; (1) 
cos (180° -A)=- sin (90° -A) = - cos (-'a) = - cos A ; (2) 
tan (180°-^)= -cot (90° - A) = tan (-4)= - tan A (3) 

Comparing the last members of (1), (2), (3) with their first 
members we have 

(i) Any trigonometric ratio of an angle is equal arithmetic- 
ally to the same ratio of its supplement; but only the sines 
(or the cosecants) of two supplementary angles have the same 
quality. 

E.g., sin 150° = sin 30°, tan 165° = - tan 15° ; 

cos 135° = - cos 45°, cot 155° = - cot 25°. 

Ex. 1. Express in terms of a trigonometric ratio of its supplement 
each trigonometric ratio of 125° ; 143° ; 157°. 

Comparing the last members of (1), (2), (3) with their 
second members we have § 9 generalized ; that is, 

(ii) Any trigonometric ratio of an angle is equal to the 
co-ratio of its complement. 

Ex. 2. Applying § 29 three times in succession and § 28 once, we have 
sin (270°±^4)ee cos(I80°±^L) =-sin (90°±^4) ee-cos(±^4)=-cos^ ; 
cos (270° ±A) =- sin (180° ±A) ee- cos (90° ±A) = sin (±A) =± sin A ; 
tan (270°±^) EE-cot (180°±^.) ~ tan (90°±A) =-cot (± A) EE=fcot A. 

Ex. 3. Prove (ii) by putting - A for A in (1) and (2) of § 29. 

31. Trigonometric ratios of n • 90 + A, in terms of the ratios 

of A. To obtain in terms of a trigonometric ratio of A any 
trigonometric ratio of n • 90° + A (where n is a positive inte- 
ger), we apply § 29 n times in succession ; and to obtain in 
terms of a ratio of A any ratio of w"-90° — A, we first apply 
§ 29 n times and then § 28 once. In each case we change 
from ratio to co-ratio n times ; hence 



TRIGONOMETRIC RATIOS OF n -90° ± A 45 

(i) When n is even, any trigonometric ratio of n • 90° ± A is 
equal arithmetically to the same ratio of A. 

(ii) When n is odd, any trigonometric ratio of n • 90° ± A is 
equal arithmetically to the co-ratio of A. 

When .4 is a positive acute angle, any trigonometric ratio 
of A is positive ; hence 

(iii) The two trigonometric ratios in (i) or (ii) will be oqypo- 
site in quality when, and only when, the ratio of n . 90 ± A is 
negative for A positive and acute. 

Any positive angle can be written in the form n • 90° ± A 
where A has some positive value less than 45°. 

E.g., 580° = 6 • 90° -f 40° ; here n is even, and the angle is in the third 
quadrant. Hence, by (i) and § 22, we have 

sin 580° - sin (6 • 90° + 40°) = - sin 40° ; 
cos 580° = cos (6 • 90° + 40°) = - cos 40° ; 
tan 580° = tan (6 • 90° + 40°) = tan 40° ; etc. 

Again, 270° + A = 3 • 90° + A ; here n is odd, and 270° + A is in the 
fourth quadrant when A < 90°. Hence, by (ii) and (iii), we have 

sin (270° + A) = - cos A, tan (270° + A) = - cot A ; 
cos (270° + A) = sin A, cot (270° + A) = - tan -4 ; etc. 

Example. Express in terms of a trigonometric ratio of A each trigo- 
nometric ratio of 180° + A ; 180° - A ; 270° - A ; 360° ± A. 

EXERCISE IX 

Express each of the following trigonometric ratios in terms of the ratio 
of some positive acute angle less than 45°. 

1. sin 168°. 6. cos (-84°). 11. cot 1054°. 

2. tan 137°. 7. tan (-246°). 12. sec 1327°. 

3. cos 287°. "8. cos (-428°). 13. esc 756°. 

4. sin 834°. 9. cos 1410°. 14. tan (- 196° 54') . 

5. sin (-65°). 10. tan 1145°. 15. cot (- 236° 21'). 



46 



PLANE TRIGONOMETRY 



16. Prove sin 420° • cos 390° + cos (- 300°) • sin (- 330°) = 1. 

17. Prove cos 570° • sin 510° - sin 330° • cos 390° = 0. 

32. Trigonometric lines representing the trigonometric ratios. 

Any trigonometric ratio is a positive or a negative number, 
but it can always be represented by a directed line, as below. 
Let A denote any angle coterminal with Z XOP in each of 
the four figures. Take OP as a positive unit line, and draw 
PM± OX. 

Then sin A = MP J OP = the numerical measure of MP ; (1) 
hence sin A is represented by the directed line MP. 

Also, cos A = OM J OP = the numerical measure of OM ; (2) 
hence cos A is represented by the directed line OM. 

T 




X<[*U 



Fig. 26 



TRIGONOMETRIC LINES 47 

To obtain directed lines which shall represent the four other 
trigonometric ratios, draw OY A. OX at 0, and take 
OX = OY = OP = a positive unit line. 

At X draw XT _1_ OX, at Y draw YC ± OY, and prolong each 
until it meets the final side OP (produced through P or 0) in 
some point as T or C. 

According to the laws assumed in § 21 for the quality of 
MP, OM, and OP, the directed line XT is positive or negative 
according as it extends upward or downward from its origin 
X\ YC is positive or negative according as it extends to the 
right or to the left from its origin F; and OT or OC is posi- 
tive or negative according as it extends from the origin in 
the direction of the final side OP or in the opposite direction. 

E.g., XT or YC is + in fig. a or c, and - in fig. b or d. 
OT is + in fig. a or d, and — in fig. b or c. 
OC is + in fig. a or 6, and — in fig. c or d. 

In each figure the triangles OMP, OXT, and OYC, being 
mutually equiangular, are similar. 

In each of the four figures we find that 
tan A = MP/OM = XT/ OX = the numerical measure of XT; (3) 
hence tan A is represented by the directed line XT. 
cot A — OM/MP = YC /OY = the numerical measure ofYC; (4) 
hence cot A is represented by the directed line YC. 
sec A = OP / OM = OT /OX = the numerical measure of OT ; (5) 
hence sec A is represented by the directed line OT. 
esc A ■= OP / MP = 0C / OY = the numerical measure of OC ; (6) 
hence esc A is represented by the directed line OC. 

The directed lines which represent the trigonometric ratios 
of an angle are called the trigonometric lines of that angle. 

The relations in (1) to (6) can be written briefly 
sin A = MP, cos A = OM, tan A = XT, 
cot A = YC, sec A = OT, esc A = OC. 



48 PLANE TRIGONOMETRY 

Since the trigonometric lines represent graphically the trigo- 
nometric ratios, or, in other words, the trigonometric ratios are 
the numbers which measure the trigonometric lines, it follows 
that if we prove any relation between the trigonometric lines, 
we know that the same relation exists between the correspond- 
ing trigonometric ratios, and vice versa. 

33. Use of trigonometric lines in proofs and discussions. To 

fix in the pupil's mind the trigonometric lines which represent 
the trigonometric ratios, to help familiarize him with the use 
of directed lines to represent positive and negative real num- 
bers, and to show him how the use of the trigonometric lines 
sometimes simplifies trigonometric proofs and discussions, we 
give below illustrative examples, which can be taken or omitted 
at the option of the teacher. 

Ex. 1. Using trigonometric lines, prove the relations in § 24. 
In each of the four figures in § 32 we have 

MP 2 + 7)M 2 = ~OP 2 , .-. sin 2 ^ + cos 2 ^.= 1; 

OX 2 + XT 2 = ~OT 2 , .-. 1 + tan 2 J. ee sec 2 .4; 

~dY 2 +YC 2 = ~OC 2 , .-. 1 + cot 2 ^4eecsc 2 ^4; 

XT/OX=MP/OM, .-. tan A = sin A /cos A ; 

YC/OY = OM/MP, .-. cot A = cos A /sin A ; 

OT/OX = 'OP/OM, .-. sec A = l/cosA; 

OC/OY- OP /ON, .-. esc A = l/smA. 

Ex. 2. Using the trigonometric lines, determine the quality of each 
trigonometric ratio in each quadrant. 

In the figures of §32, the quality of MP, or sin ^4, is easily deter- 
mined. So also is the quality of OM, or cos A. 

XT, or tan A, is positive when A is in the first or the third quadrant, 
and negative when A is in the second or the fourth quadrant. 

OT, or sec A, extends in the direction of OP, and is therefore positive 
when A is in the first or the fourth quadrant ; and OT, or sec A, extends 
in the direction opposite to that of OP, and is therefore negative when A 
is in the second or the third quadrant. 

In like manner determine the quality of YC, or cot ^4, and of OC, or 
csc^L. 



TRIGONOMETRIC LINES 



49 



Ex. 3. Using trigonometric lines, trace the changes of the trigono- 
metric ratios of A while A increases from 0° to 360°. 

In the figures of §32, the changes of MP, or sin^l, and of OM, or 
cos^i, are easily followed. 

While A increases from 0° to 90° (fig. a), XT beginning at zero increases 
without limit as A approaches 90° ; i.e. tan A increases from to + oo. 

While A increases from 90° to 180° (fig. b), XT is at first of infinite 
length and negative, and becomes when A = 180° ; i.e. tan A increases 
from — oo to 0. 

While A increases from 180° to 270° (fig. c), XT beginning at zero 
increases without limit as A approaches 270°; i.e. tan A increases from 
to + oo. 

While A increases from 270° to 360° (fig. d), XT is at first of infinite 
length and negative, and becomes when A — 360° ; i.e. tan A increases 
from — oo to 0. 

In like manner the student should trace the changes of cot A, sec A, 
and esc A. 



Ex. 4. Using the trigonometric lines, find the trigonometric 
180° - A and 180° + A in terms of 
those of A, when A is in the first C^_ 
quadrant. 

Let the angles XOP, WOP', 
and WOP" be equal in size ; then 
if A is coterminal with XOP, 
180° — A will be coterminal with 
XOP', and 180° + A will be coter- 
minal with XOP". 

Draw the trigonometric lines of 
A, 180° - A, and 180° + A. 



ratios of 
C 




Fig. 27 



Then 



Again, 



M'P' = MP, 
OM' = - OM, 

XT' = - XT, 

YC = - YC, 

OT'=- or, 

OC'= OC, 

M"P" = - MP, 

OM" = - OM, 
XT = XT, 
YC = YC, 



.\ sin (180° - A) = sin A; 
.-. cos (180° - A) = - cos A 
.-. tan (180° - A) = - tan A 
.-. cot (180° - A) ~ - cot A 
.-. sec (180° -A)= -sec^l 
.-. esc (180° - A) = csc A. 
.-. sin (180° + ^4) = - sin A ; 
.-. cos (180° + A) = - cos A ; 
.-. tan (180° + A) = ton A; 
.-. cot (180° + ^4) = cot^. 



50 PLANE TRIGONOMETRY 

OT and OC are both negative when the angle is XOP". 

Hence sec (180° + A) = - sec A, esc (180° + A) = - esc A. 

A similar proof could be given when A is in any one of the other 
three quadrants. 

Ex. 5. Using the trigonometric lines, find the trigonometric ratios of 
— A in terms of those of A. 
In each figure of § 28, let 

OP — OP' = a positive unit line. 
Then MP = sin A , OM = cosA, 

M'P' = sin (- A), OM' = cos (- A). 

But in each figure we have 

M'P' = - .MP, OJr = Oiif . 

Hence sin (— A) = — sin J., cos (— J.) = cos^i. (1) 

From the identities (1) we can obtain the other relations as in § 28. 

The student should draw the other trigonometric lines of A and — A 
in each of the four figures in § 28, and prove the last four identities by 
the use of these lines. 

Ex. 6. Using the trigonometric lines, find the trigonometric ratios of 
90° + A in terms of those of A. 
In each figure of § 29, let 

OP — OP' — a positive unit line. 
Then MP = sin A , OM =cosA, 

M'P' = sin (90° + A), OM' = cos (90° + A). 
But in each figure we have 

M'P' = OM, OM' = - MP. 

Hence sin (90° + A) = cos A, cos (90° + A) = - sin A. (1) 

From the identities (1) we can obtain the other relations as in § 29. 

But the student should draw the other trigonometric lines of A and 
90° 4- A in each of the four figures in § 29, and prove the last four identi- 
ties by the use of these lines. 

Ex. 7. Using the trigonometric lines, find the trigonometric ratios of 
90° — A in terms of those of A, when A is in the first quadrant. 



TRIGONOMETRIC LINES 51 

In the figures of § 32, let X be the origin of arcs, and let the 
arc be positive or negative according as its generating point 
moves counter-clockwise or clockwise. Then the arc in each 
figure will have the same numerical value in degrees as the 
angle which it subtends at the center, and the trigonometric 
lines of the angle in each figure can be regarded as trigono- 
metric lines of the arc, and the ratios which these lines repre- 
sent can be regarded as trigonometric ratios of these arcs. 
Hence, if the number of degrees in a unit arc is equal to the 
number of degrees in an angle, the arc and the angle have the 
same trigonometric lines or ratios. 



CHAPTEE III 
TRIGONOMETRIC RATIOS OF TWO ANGLES 

34. Sine and cosine of the sum of two angles. Let XOR and 

ROC be any two positive acute angles. 

Then Z XOR + Z ROC = Z XOC. 

Let A denote any angle, positive or negative, coterminal 
with Z XOR, and B any angle coterminal with Z ROC. 
Then the sum A + B will be coterminal with Z XOC. 








>X 



Fig. 28 



The sum A + B may be in the first quadrant, as in fig. a, or 
in the second quadrant, as in fig. b. 

In each figure, from any point on OC, as P, draw PN _L Oi2 
and PM _L OX ; also draw NQ _L OX, and ND _L MP. 

Then the triangles DPN and Q6W will be similar. 

Now MP = QN + DP. 

By § 21, QiVEE sin ,4- ON. 

Again, DP /NP = OQ/ ON = cos .4 ; 

whence DP = cos J. • iVP. 

52 



ADDITION FORMULAS 53 

Hence MP = sin A • ON + cos A • A 7 P. 

.'. MP J OP = sin .4 • OZVy OP + cos ^ • NP / OP. (1) 

Substituting for the ratios in (1) their names, we have 

sin (A + B) = sin A cos B + cos A sin B. [7] 

Again, OM=OQ- DN 

= cos A • ON — sin A • NP. 

.'. OM/ OP = cos 4 • 02V/ OP - sin A • NP / OP. 

. ' . cos (A + B) = cos A cos B — sin A sin B. [8] 

Observe that thus far [7] and [8] are proved only when the angles 
A and B are both in the first quadrant. In § 35 it will be shown that 
these relations hold true in whatever quadrant A or B is. 

35. General proof of [7] and [8]. 



sin (.4 + 90° + B) = sin (90° + A + B) 

eecos(.4 +B) §29 

= cos A cos B + (— sin .4) sin B by [8] 

= sin (A + 90°) cos B + cos (.4 + 90°) sin B. (1) 



Again, cos (A + 90° + B) = cos (90° + A + B) 

= - sin (.4 ±B) § 29 

= (— sin .4) cos B — cos A sin B by [7] 

= cos (.4 + 90°) cos B - sin (.4 + 90°) sin B. (2) 

Now in whatever quadrant A is, .4 + 90° is in the next 
quadrant. Hence, from (1) and (2), it follows that if [7] and 
[8] are true when A is in any one quadrant, they are true 
also when A is in the next quadrant. But, by § 34, [7] and 
[8] are true when A is in the first quadrant ; hence they are 
true when A is in the second quadrant; and so on. Hence 
[7] and [8] hold true in whatever quadrant A is. 

The same reasoning applies to B. Hence [7] and [8] hold 
true for all values of A and B, positive or negative. 



54 PLANE TRIGONOMETRY 

Formulas [7] and [8], often called the 'addition formulas, are 
very important and should be memorized. 

So many theorems can be deduced from the formulas [7] and [8] that 
they are often called the fundamental formulas of trigonometry. 



EXERCISE X 

1. State in words identities [7] and [8], as generalized in § 35. 

The sine of the sum T _ ( sin first • cos second 
of any two angles j — ^ -f cos first • sin second. 

2. sin 75° = sin (30° + 45°) 

= sin 30° cos 45° + cos 30° sin 45° by [7] 

_ 1 V2 V3 V2 _ V2 -f V6 
~ 2*~2~ + ~2 2~ ~ 4 

3. Putting 75° = 30° + 45° and using [8], find cos 75°. 

4. Putting 15° = 45° + (- 30°), find sin 15° and cos 15°. 

5. Putting 15° = 60° + (- 45°), find sin 15° and cos 15°. 

6. Putting 90° = 60° + 30°, find sin 90° and cos 90°. 

7. Putting 0° = 45° + (- 45°), find sin 0° and cos 0°. 

A and B being positive acute angles, find the values of sin [A + B) 
and cos ( A + B), having given 

8. sin A = 2/5, cos B = 1/3. 9. sin A = 2/3, cos B = 1/4. 

10. Putting 90° + A for A in [7], deduce [8]. 

11. Putting 90° + A for A in [8], deduce [7]. 

12. Prove [7] and [8], using trigonometric lines, A and B being in 
the first quadrant. 

Take OP = + 1. 

Then MP = sin (A + 5),. 

ON = cos B, NP = sin B. 
.-. DP = NP cos DPN = sin B cos A. 
QN = ON sin A = cos B sin A . 
.-. sin (A -\- B) - QN + DP = sin A cos B -f cos A sin B. 



SUBTRACTION FORMULAS 55 

36. Sine and cosine of the difference of two angles. Substitut- 
ing — B for B in [7], we have 

sin (A — B) = sin A cos (— B) + cos A sin (— B). 
.'.sin (A — B) = sin A cos B — cos A sin B. [9] 

Substituting — B for B in [8], we have 

cos (A — B) = cos .4 cos (— B) — sin A sin (— B). 
.'.cos (A — B) = cos A cos B -f sin A sin B. [10] 

Formulas [9] and [10] are often called the subtraction 
formulas. 

EXERCISE XI 

1. State in words identities [9] and [10]. 

The sine of the difference ~) _ f sin first • cos second 
of any two angles J ~ I ~~ cos first • sin second. 

2. Putting 15° = 45° - 30°, find sin 15° by [9] and cos 15° by [10]. 

3. Putting 15° = 60° - 45°, find sin 15° and cos 15°. 

A and B being positive acute angles, find the values of sin (A — B) and 
cos {A — jE>), having given 

4. sin A = 1/4, sin B= 1/3. 5. cos A = 2/3, cos B- 3/4. 

Prove each of the following identities : 

6. sin (A + B) sin (A - B) = sin 2 A cos- B - cos 2 A sin 2 B 

+ (sin 2 A sin 2 B - sin 2 A sin 2 B) 
= sin 2 J. (cos 2 B + sin 2 B) - sin 2 B(cos 2 A + sin 2 ^l) 
= sin 2 ^ -sin 2 B. 

Observe that sin 2 Asm 2 B — sin 2 A sin 2 5 is added above as one form 
of zero. 

7. cos (A + B) cos ( J. - B) = cos 2 ^4 - sin 2 B. 

8. sin (^1 + B) cos 5 — cos (^4 + B) sin B = sin ^4. 

9. sin (A + B) + cos ( J. - B) = (sin J. + cos ^4) (sin B + cos B). 
10. sin A cos (J5 - C) - sin B cos ( A + C) = sin (A - B) cos C. 




56 PLANE TRIGONOMETRY 

11. tan A -f tan B = sin (A + B)/(cos A cos B). 

12. cot i? - cot A = sin (J. - 5)/(sin J. sin B). 

13. Prove [9] and [10] geometrically, using trigonometric lines, when 
A, B, and A — B are in the first quadrant. 

Let XOR and ROC be any two acute angles, ZROC 
being negative and Z XOC being positive. 
Then Z XOC = Z XOR + Z ROC. 

Let A denote any angle coterminal with XOR, and 
— B any angle coterminal with ROC. 
^X Then A + (— B), or J. — 5, will be coterminal 
with XOC. 

Take OP equal to -f 1. 
Draw PM _L OX, PN ± OR, NQ ± OX, and PB JL QJ\T. 

Now NP = sin (- 5) = - sin 5, CW= cos ( - 5) = cos B, 

and sin (A - B) = MP = QN - DN. 

Also, QN= ON- sin XOE = cos 5 sin A, 

and Z>iV = ( - NP) cos DiVP = sin BcosA. 

.-. sin (^4 — B) = QN — DN — sin A cos B — cos ^4 sin B. 
Again, OQ — ON cos XOR = cos 5 cos A, 

and DP = (- NP) sin DiVP = sin J5 sin A. 

.-. cos (J. - B) = OM = OQ + DP = cos A cos B + sin ^L sin 5. 

37. Tangent of the sum and difference of two angles. Divide 
the members of [7] by those of [8] ; then by [2] we have 

sin A cos B + cos A sin B 

tan {A + B) = p . . . • (1) 

v 7 cos A cos i? — sin A sin 5 

To express tan (J. + 5) in terms of tan A and tan B we 
divide the numerator and denominator of the fraction in (1) 
by cos A cos B ; then by formula [2] we obtain 

«v tan A + tan B _ . . _ 

tan (A + B) = — A -• [111 

v T ' 1 - tan A tan B L J 

Substituting — B for B in [11], we obtain 

^n tan A — tan B _.__ 

tan (A — B) = -• [121 

v ' 1 + tan A tan B L J 



TRIGONOMETRIC RATIOS OF 2 A 57 

EXERCISE XII 

1. State in words identities [11] and [12]. 

The tangent of the sum "1 _ f the sum of their tangents 
of any two angles J _ 1 1 — product of their tangents 

2. Putting 75° = 45° + 30°, find tan 75° by [11]. 

3. Putting 15° = 60° - 45°, find tan 15° by [12]. 

4. If tan A = - 1/2 and tan B = 3, find tan (A + B) and tan (A - B). 

5. If tan A = - 2 and tan B = - 3, find tan (A + B) and tan (A - B). 
Prove each of the following identities : 

n * tAtio . a\ 1+tan^L _cot A cot B - 1 

6. tan (45° + ^) = 8. cot(A + B) = 

1 - tan A cot B + cot A 

r, < ,Arn a* I— tan A cotAcotB + 1 

7. tan(45°-^L) = 9. cot(A-B) = 

1 + tan^l cot B - cot A 

10. Prove identity [12] by dividing [9] by [10]. 

11. Prove the identities in examples 8 and 9 by taking the reciprocals 
of the members of [11] and [12] respectively. 

12. Find tan (A -f B) and tan (A — B) in terms of cot A and cot B. 

13. Find cot (A + B) and cot (A — B) in terms of tan A and tan B. 

38. Trigonometric ratios of twice an angle in terms of the ratios 

of the angle. Substituting A for B in [7], we have 

sin (.4 + A) = sin A cos A + cos A sin A ; 

that is sin 2 A = 2 sin A cos A. [13] 

Substituting A for B in [8], we obtain 

cos 2 A = cos 2 A — sin 2 A (i) 

= 1-2 sin 2 A (ii) [ [14] 

= 2 cos 2 A - 1. (iii) 

To derive (ii) or (iii) from (i), we use identity [4]. 

Substituting A for B in [11], we obtain 

^ . 2 tan A r . _ 

tan2 A eee — . [15] 

1 - tan 2 A L J 



58 PLANE TRIGONOMETRY 

EXERCISE XIII 

1. State in words identities [13], [14], and [15]. 

sin twice an angle = 2 sin angle • cos angle. 
cos twice an angle = (cos angle) 2 — (sin angle) 2 . 

2. From the trigonometric ratios of 30°, find sin 60°, cos 60°, tan 60°. 

3. From the trigonometric ratios of 60°, find sin 120°, cos 120°, 
tan 120°. 

4. Express sin 6 A, cos 6 ^4, tan 6 A in terms of the trigonometric 
ratios of 3 A. 

5. Express sin 3 A, cos 3 A, tan 3 A in terms of the trigonometric 
ratios of 3^4/2. 

Prove each of the following identities : 

n . n a COt2 -4-1 o • o a 1-COS2J. 

6. cot 2 A = 8. sm 2 A = 

2 cot A 2 

7. esc 2 A = (sec A esc A)/ 2. 9. cos 2 ^4 = (1 + cos 2 A)/2. 

a A sec 2 A 1 + tan 2 ^4 

10. sec 2 A = = 

2 -sec 2 ^4 l-tan 2 JL 

11. cos 4 J. = 2 cos 2 2 J. - 1 ee 2 (1-2 sin 2 .A) 2 - 1 

= 8 sin 4 A -8 sin 2 A + 1. 

12. sin 4 ^4. = 4 sin ^4 cos J. — 8 sin 3 ^4 cos ^4. 

39. Trigonometric ratios of half an angle in terms of the cosine 
of the angle. Solving (ii) and (iii) of [14] for sin 2 A and 
cos 2 A respectively and putting A/2 for A, we obtain 



# A _ /l — cos A 

. A /l + cos A ,_„_ 

and cos- = ^-3_ [17] 

Divide [16] by [17], tan^ = ^~^ . [18] 



TRIGONOMETRIC RATIOS OF A/2 59 



EXERCISE XIV 

1. State in words identities [16], [17], and [18]. 

, 7 - , , 1 — cos angle 

sin half an angle = square root of — . [16] 

2. Find sin 22J°, cos 22|°, tan 22|°, from cos 45°. 
cos 45° _ / 1 - V2/2 V2-V2 



sin 221° = 



^V L 



2 \ 2 2 

3. Find sin 15°, cos 15°, tan 15°, from cos 30°. 

4. cos A = 1 /S ; find the sine, cosine, and tangent of A/ 2. 

5. cos A = a ; find the sine, cosine, and tangent of J./2. 

6. Express sin A, cos -4, and tan A in terms of cos 2 -4. 

7. Express sin 2 -4, cos 2 4L, and tan 2 A in terms of cos 4 -4. 

8. Express sin 3 A, cos 3 A, and tan 3 4. in terms of cos 6 -4. 
Prove each of the following identities : 

f2 A _ 1 + cos4._ / sin A \ 2 _ /l -f cos,4\ a 
2 — 1 — cos A ~ \ 1 — cos A ) \ sin 4. / 

1A 9 -4 / sin4. \ 2 /l-cos4\ 2 

10. tan 2 — = ( ) = ( ) = (esc A — cot A) 2 . 

2 \l + cosA/ \ sin A / v ' 

n A 2 sec J. , rt _ 4. 2 sec J. 

11. sec 2 — = 12. esc 2 — = - 



2 sec -4 + 1 2 sec A — 1 

13. Express cos 4 4. in terms of cos 2 4. and cos 4 .4. 

(cos 2 J.) 2 = (J + J cos 2 ^) 2 

= J + i cos 2 4. + J cos 2 2 4. 

= J + J cos 2 4. + i(i + J cos 4 4.). 
.-. cos 4 -4 = J + J cos 2-4 + -J- cos 4 .4. 

14. Prove sin 4 A = | — J cos 2 4. + i cos 4 4.. 

15. Prove sin 2 4. cos 2 4L = \ — icos 4 4. 
Suggestion, sin 2 4. cos 2 A = (sin 4 cos A) 2 = \ sin 2 2 A. 

16. Prove sin 2 4. cos 4 A = y 1 -^ + \ sin 2 2 4. • cos 2 4. — T ^ cos 4 -4. 
Suggestion, sin 2 ^1 cos 4 ^4 = (sin A cos 4) 2 • cos 2 A. 



60 PLANE TRIGONOMETRY 

40. Sum and difference of sines and cosines. Adding and 
subtracting [7] and [9], and [8] and [10], we obtain 

sin (A+B) + sin (.4 — B) = 2 sin A cos B ; (1) 

sin (A + B) — sin (A — B) = 2 cos 4 sin 5 ; (2) 

cos (A+B) + cos (4 — £) = 2 cos ^ cos B ; (3) 

cos (4 + B) — cos (J. — B) ~ — 2 sin A sin 5. (4) 

Let A + B = C and A - B = D. ^ 

Then A = (C + 2))/ 2, J5 = (C - Z>)/2.| 

Substituting in (1) • • • (4) the values in (5), we obtain 

C + D C — D 

sin C + sin D = 2 sin cos -. [19] 

z z 

C + D C — D 

sin C — sin D = 2 cos sin [20] 

z z 

C + D C — D 

cos C + cos D = 2 cos cos T21] 

T 2 2 L J 

C + D C — D 

cos C — cos D = — 2 sin sin [22] 



By formulas [19] • • • [22], a sum or a difference of the sines 
or the cosines of two angles is transformed into a product. 
Hence these formulas, often called product formulas, are useful 
in adapting other formulas to the use of logarithms. 

E.g., sin 7 -A + sin 6 J. = 2 sin £(7 J. + 5 A) cos \{1 A - 5^4) 

= 2 sin 6 A cos A . 
.-. log (sin 7 A -f sin 5 A) = log 2 + log sin 6 A + log cos A. 
Again, cos 8 A - cos 2 J. = - 2 sin | (8 A + 2 J.) sin J (8 J. - 2 A) 

= — 2 sin 5 A sin 3 ^L. 

By the converses of formulas (1) • • • (4), & product involving 
sines or cosines or both is transformed into a sum or a differ- 
ence of sines or cosines. 



IDENTITIES 61 



EXERCISE XV 

1. State in words identities [19] • • • [22]. 

The sum of the sines ^ _ . _ _. .»„,.- 

- . , y = 2 sm ZiaZ/ sum • cos half difference, 

of any two angles j 

Prove each of the following identities : 

2. sin 60° -f sin 30° = 2 sin 45° cos 15°. by [19] 

3. sin 50° + sin 10° = 2 sin 30° cos 20°. 

4. cos 75° + cos 15° = 2 cos 45° cos 30°. by [21] 

5. cos 80° - cos 20° = - 2 sin 50° sin 30°. 

6. sin 7 A — sin 3 A = 2 cos 5 A sin 2 A. 

„ sin 7 A — sin 5 A 2 cos 6 Asm A 

7. = = tan^4. 

cos 7 A + cos 5 ^4 2 cos 6 J. cos ^4 

8. s^ + sinS^^^ 
cos A + cos 3 A 

. sin C + sin D C + D C - D tan £ (C + D) 

9. = tan cot = ~ -. 

sin C — sin D 2 2 tan £ (C — D) 

,. sin C + sin B C + D 
10. = tan 

cos O + cos D 2 



11. 
12. 
13. 



sin C + sin X) 

cos C — cos D 

sin C — sin D 

cos C -f cos D 

sin O — sin D 
cos O — cos D 




, - cos C + cos D 

14. = — cot cot - 

cos C - cos D 2 2 

ee cot | (C + D) cot -j-(D- C). 

15. Given sin A = 1 /2, sin B = 1 /3, to find sin (J. + 5), sin (^4 - £)< 

cos (A + B), cos(J.-J5), sin 2 ^4, sin 2 2?, cos 2 ^4, cos 2^: (1) when A 
and 5 are both in the first quadrant ; (2) when A is in the first and B is 
in the second quadrant. 



62 PLANE TRIGONOMETRY 

16. From the answers to example 15, find in the simplest way 
tan (A + B), tan (^1 - B), cot (A + B), cot {A - B), sec (A + B), 
esc (A 4- B), tan 2 J., cot 2 .A, sec 2 5, esc 2 B, in cases (1) and (2). 

EXERCISE XVI 
Examples for Review 

Prove each of the following identities : 

„ sin (x 4- y) tan x + tan y „ n . 2 — sec 2 A. 

1. * zi = *. 5. cos 2 A. = 

sin (x — y) tan x — tan y sec 2 J. 

rt cos (x + y) 1 — tan x tan y n rt . esc 2 A 

= 6. sec 2 J. f= 



cos (x — y) 1 4- tan x tan 2/ esc 2 A. — 2 

cos (x + y) „ . _ . 2 tan A 

• = cot x — tan y. 7. sin 2i = 



sin x cos 2/ 1 + tan 2 A 

. cos (x - y) , , _ sin 3 J. - sin J. 

4. — = tan x + cot y. 8. = tan A. 

cos x sin y cos 3 ^i 4- cos A 

9. Express sin (3 x/ 4), cos (3 x/ 4), and tan (3 x/ 4) in terms of 
cos(3x/2). 

10. Express sin (3 x/ 4), cos (3 x/ 4), and tan (3 x/ 4) in terms of the 
trigonometric ratios of 3x/8. 

Prove each of the following identities : 

11. (sin J. + cos A) 2 = 1 +sin2A. 

12. (sin A — cos A)' 2 = 1 — sin 2 A. 

13. tan ^i -f- cot A = 2 esc 2 J.. 

14. cot J. — tan A = 2 cot 2 A. 

, K tan J. + tan 5 . A _ 

15. = tan A tan 5. 

cot ^1 + cot B 

16. Given sin^i =2/3, cos £=1/2, to find (1) sin (-44-5), sm(A-B), 
cos(A + B), cos(A-B), sin2^i, cos2J., sin2£, cos2£; (2) tan (A 4- B) , 
cot (4 4- B), tan (4 - B), cot (J. - £), tan 2 ^L, cot 2 A, tan 2 5, cot 2 jB. 

Prove each of the following identities : 

- „ cot J. + cot J5 . _ . . . _ . , 

17. = cos (B - A) sec (B 4- A). 

cot A - cot B v ' v ' 



IDENTITIES 63 

18 rin^ + J)sin(A-B) stang ^_ tim , jB , 

cos 2 ^i cos 2 B 

19. tan 2 ^-tan 2 ^^ ta ^ ^ 
l-tan 2 ^itan 2 £ \ > \ J 

20. V2 sin (J. ± 45°) = sin A ± cos A. 

21. 2 sin (45° - A) cos (45° + B) = cos (^L - B) - sin (^1 + B). 

22. 2 sin (45° + A) cos (45° + B) = cos (i 4- 5) + sin (4 — 5). 

23. 2 sin (45° 4 ^4) cos (45° - B) = cos (J. - £) 4 sin {A + 5). 
cot A — 1 /l — sin 2 ^i 1 — sin 2 ^1 



p.nt. /4 

25. cot (.4 - 45°) = 



cot ^L 4- 1 \ 1 + sin 2 ^4 cos 2 ^1 
cot A + 1 _ tan ^4 4 1 



1 — cot A tan J. 

26. tan (A ± 45°) 4 cot (A T 45°) =0. 

27. sin 9x — sin 7x = 2 cos 8 x sin x. 

28. cos 7 x 4- cos 5 x = 2 cos 6 x cos x. 

_ _ sin 3 x - sin x . _ 

29. = cot2x. 

cos x — cos 3 x 

_ . sin 5 x — sin 2 x 7x 

30. = cot 

cos 2 x — cos 5 x 2 

sin A 4 sin 5 _ cos J. + cos JB 
cos A — cos 5 sin B — sin ^4 

32. tan (x/ 2 4 45°) = tan x 4 sec x. 



CHAPTEE IV 
SOLUTION OF RIGHT TRIANGLES WITH LOGARITHMS 

41. In Chapter I, right triangles were solved without loga- 
rithms. In general, however, arithmetic computations are 
much abbreviated by using logarithms. It is assumed that 
the student is already familiar with the theory of logarithms 
from the study of Algebra ; but to bring to mind those proper- 
ties of logarithms which adapt them to shortening arithmetic 
computations, a brief review is given below. 

42. Logarithms. If a = N, (1) 

then x, the exponent of a, is called the logarithm of N to the 
base a, which is written in symbols 

x = log a N. (2) 

Equations (1) and (2) are equivalent ; (2) is the logarithmic 
form of writing the relation between a, x, and N given in (1). 

E.g., since 3 2 = 9, 2 is the logarithm of 9 to the base 3 ; i.e. + 2 = log 3 9. 
Since 2~~ 3 = 1/8, -3 = log 2 (1/8). 

Since 4 3/2 = 8, +3/2 = log 4 8. 

Ex. 1. Express in the logarithmic form each of the following relations : 

3 4 = 81, 4 3 = 64, 6 3 = 216, n c = 6, 5~ 3 = 1/125, 3 -5 = 1/243. 
Ex. 2. Express in the exponential form each of the following relations : 
log 5 125 = 3, loga 32 = 5, log 4 64 = 3, 

log c Jlf=&, log 2 (l/16) = -4. 

Ex. 3. When the base is 10, what is the logarithm of 1 ? 10 ? 100 ? 
1000? 10000? 100000? 0.1? 0.01? 0.001? 0.0001? 0.00001? 

x Ex. 4. What is the number when the base is 10 and the logarithm 
is0?l?2?3?-l?-2?-3?-4? 

64 



PROPERTIES OF LOGARITHMS 65 

43. Properties of logarithms. Since logarithms are exponents, 
from the general laws of exponents we obtain the following 
general properties of logarithms to any base. 

(i) The logarithm of the product of tivo or more arithmetic 
numbers is equal to the sum of the logarithms of the factors. 

Let M = a x , N - av. 

Then M x N = a x +v. 

Hence log a (M + N) = x + y = log a M + log a iV. 

(ii) The logarithm of the quotient of two arithmetic numbers 
is equal to the logarithm of the dividend minus the logarithm 
of the divisor. 

Let M= a x , N = aJL 

Then M -r- N = a*-*. 

Hence log a (M -f- N) = x — y = log a If — log^. 

(iii) The logarithm of any power of an arithmetic number is 
equal to the logarithm of the number multiplied by the exponent 
of the power. 

Let M = a x . 

Then, for all real values of _p, we have 
M p = ap*. 

Hence log a (M p ) = px = p log a Jf. (1) 

If p = l/r, from (1) it follows that 

(iv) The logarithm of any root of an arithmetic number is 
equal to the logarithm of the number divided by the index of the 
root. 

An expression is said to be adapted to logarithmic computa- 
tion when it involves only products, quotients, powers, or roots. 

E.g., x c y l/r /z s is adapted to logarithmic computation ; for we have 

10g a (X c y*/ r /Z*) = C \0gaX + (l/r) loga V ~ 8 loga Z. (1) 

Observe that only the arithmetic value of a product, quotient, power, 
or root is obtained by logarithms ; the quality must be determined by the 
laws of quality. 



66 PLANE TRIGONOMETRY 

Logarithms do not aid in the operation of addition or of subtraction. 
But when, as in formulas [19] • • • [22], a sum or a difference is identical 
with a product, the sum or difference can be obtained by computing the 
product. 

E.g. , log a (x 2 - 2/ 2 ) = log a [(x + y) (x - y)] = log a (x + y) + log a (x-y). 

44. Common logarithms. The logarithms used for abridg- 
ing arithmetic computations are those to the base 10 ; for this 
reason logarithms to the base 10 are called common logarithms. 

Thus the common logarithm of a number answers the ques- 
tion, What power of 10 is the number? 

Most numbers are incommensurable powers of 10 ; hence 
most common logarithms are incommensurable numbers, whose 
approximate values we usually express decimally. 

E.g., the common logarithm 
of any number between 10 and 100 lies between +1 and +2 ; 
of any number between 1 and 10 lies between and +1 ; 
of any number between 0. 1 and 1 lies between — 1 and ; 
of any number between 0.01 and 0.1 lies between — 2 and -1 ; etc. 

Hence the common logarithm 
of any number between 10 and 100 is +1 + a positive decimal ; 
of any number between 1 and 10 is -f a positive decimal ; 
of any number between 0. 1 and 1 is -1 + a positive decimal ; 
of any number between 0.01 and 0.1 is — 2 -f a, positive decimal. 

45. Characteristic and mantissa. A logarithm is said to be 
in the type form when it is expressed as the sum of an integer ■, 
positive or negative, and a positive decimal fraction ; in this 
form the integer is called the characteristic, and the fraction 
the mantissa. 

In the following pages, when no base is written the base 10 
is understood. 

A negative characteristic, as ~~1, is usually written in the 
form 1 or 9 — 10; ~2 in the form 2 or 8 — 10; etc. 



CHARACTERISTICS 67 

The second form, which is usually the more convenient for 
negative characteristics, is sometimes used even when the 
characteristic is positive. 

E.g., log 434.1 = 2.63759 ; +2 is the characteristic and +.63759 is the 
mantissa; log 0.0769 = 2.88593, or 8.88593 - 10 ; ~2, or 8 - 10, is the 
characteristic, and +.88593 is the mantissa. 

In the first form of writing a negative characteristic, the sign — is 
written above the characteristic to show that this sign affects the char- 
acteristic only. One practical advantage of the second form is that we 
can make the positive part of any logarithm as large as we please, or the 
negative part any multiple of 10 we please. 

E.g., Iog0.0769 = 2.88593 = 8.88593-10 = 18.88593-20=.... 
Also, log 434.1 = 2.63759 = 12.63759 - 10 = 22.63759 - 20 = • • -. 

46. The characteristic of the common logarithm of a number 
is found by the following simple rule : 

Calling units* place the zeroth place, if the first significant 
figure in any number M is in the nth place, then the character- 
istic of log M is + n or — n, according as this first figure is to 
the left or to the right of units' place. 

E.g., when the first significant figure of a number, as 5348, is in the 
third place to the left of units' place, then the number lies between 10 3 
and 10 4 ; hence its common logarithm is +3 + a mantissa. 

Again, when the first significant figure of a number, as 0.00071, is in 
the fourth place to the right of units' place, then the number lies between 
10- 4 and 10— 3 ; hence its common logarithm is — 4 + a mantissa. 

Let the first significant figure in the number M be in the 
nth place to the left of units' place ; then M lies between 10 n 
and 10* +1 ; that is, 

71 r __ i f\?i + a positive decimal. 

.*. log M = + n + a mantissa. 
Again, let the first significant figure in the number M be in 
the nth. place to the right of units' place ; then M lies between 
10-" and lO-*— 1 *; that is, 

1 T __ i A — n -f- a positive decimal. 

.\ log M = -n + a mantissa. 



68 PLANE TRIGONOMETRY 

47. If the expressions of two numbers differ only in the posi- 
tion of the decimal point, the two numbers have the same mantissa. 

When, in the expression of a number, a change is made in 
the position of the decimal point, the number is multiplied or 
divided by some entire power of 10 ; that is, an integer is 
added to or subtracted from its logarithm ; therefore its man- 
tissa is not changed. 

E.g., 34.271 x 10 3 = 34271. 

.-. log 34.271 + 3 = log 34271. § 43, (i) 

Hence the mantissa for 34.271 equals the mantissa for 34271. 

48. A convenient formula for computing the common logarithms 

of whole numbers is 

log( g + l) S log» + 2 W i(^ T + 3(2g 1 +1)t + -) (a) 

where m = 0.434294, and z is any whole number. 

For the proof of identity (a) see § 97 in Taylor's Calculus 
or § 322 in Taylor's College Algebra. 

To compute log 2, put z = 1 in (a) ; to compute log 3, put 
z = 2 ; to compute log 4, we have log 4 = 2 log 2 ; to compute 
log 5, put z = 4 ; and so on. 

The series in (a) converges more and more rapidly as z 
increases. 

Note. Before proceeding farther in this chapter, the student should 
familiarize himself with the use of logarithmic tables, both of natural 
B numbers and of the trigonometric ratios of 
angles. 

An explanation of the tables will be found 
in the introduction to them. 

49. Right-angled triangles. Review 
§13. 

Case (i). Ex. 1. In the right triangle 
ABC, A = 48° 17', and AB = 324 ft.; solve 
Fig. 30 the triangle. 




RIGHT-ANGLED TRIANGLES 69 

rJ3 = 41 43', 

Given < A = tT A 17 '' to find J a = 241.85, 

i c = 324 ; 
L [ 6 = 215.6. 

Construct the triangle ABC, having the given parts. 

f B = 90' - A = 41° 43 / , 
Formulas < a = c sin A, 

[_ b = c cos -4. 

f log a = log c + log sin ^4, 
Logarithmic formulas < , , . , . . 

L log 6 = log c + log COS -4. 

log c = 2. 51055 log c = 2. 51055 

log sin A = 9.87300 - 10 log cos A = 9.82311 - 10 

.-. log a = 2.38355 .-. log b = 2.33366 

.-. a = 241.85. .-.6 = 215.6. 

In Chapter I we checked, or verified, the calculated values by construc- 
tion and measurement. But these values are more usually checked, or 
tested, by using some known relation between the sides and angles which 
has not been employed in solving the triangle. Thus, in the example 
above, we might use either the relation a 2 = c 2 — 6 s or tan A = a/ 6 as a 
check ; but the former is the better. 

Check. a* = (c + 6) (c - 6). log (c + 6) = 2.73207 

Here c + 6 = 539.6, lo S (c - &) = 2.03503 

c - 6 = 108.4. ••• lo S a = 4.76710/2 

= 2.38355. 

As this value of log a is the same as that obtained in the solution above, 
the answers are probably correct to four figures. 

Before using the tables the student should make a complete 
outline of the computation (such as he would have by erasing 
the second members of the equations following the logarithmic 
formulas). 

Note 1. The direction above enables the student to save time by 
writing at once all the logarithms that are found at one place in the 
table. Thus we find log sin A and log cos A at the same time ; then 
having both log a and log 6, we next find a and 6. 

Note 2. When the student has become familiar with logarithmic 
computations, he need not write the logarithmic formulas. By a glance 



70 



PLANE TRIGONOMETRY 



at the trigonometric formulas he will know how to combine the logarithms 
in the computation and can arrange his work accordingly. 

Note 3. As a check formula, we use a 2 = (c -f 6) (c — 6) or b 2 = 
(c + a) (c — a), according as c — b or c — a is the greater. 

Case (ii). Ex. 2. In the right triangle ABC, AB 
= 18.7 ft. and CB = 16.98 ft.; solve the triangle. 

'A = 65°W, 
B = 24° 46', 
6 = 7.8339. 




Given 



{.: 



= 18.7, 
= 16.98; 



to find 



Fig. 31 



Construct the triangle ABC, having the given parts. 

'sin ^4 = a/c, 

B = 90° -A, 
b = a cot A. 



.■■-.-. , f log sin A = log a — log c, 

Logarithmic formulas i , , . , , . . 

& ^ log b = log a + log cot A. 



log a: 

l0gC: 



11.22994 - 10 
1.27184 



log sin A = 9.95810-10 

.-. A = 65° 14', .-. B = 24° 46 



loga = 1.22994 
log cot A = 9.66404 - 10 
.-.log b = 0.89398 
.-. b = 7.834. 



Check. a 2 = (c + b) (c-b). 



Here 



c + 6 = 26.534, 
c - b = 10.866. 



log (c + b) = 1.42380 
log (c - b) = 1.03607 
.-.log a = 2.45987/2 
= 1.22994. 



As this value of log a is the same as that obtained from the table, the 
answers are probably correct to four places. 



Ex. 3. 



fa = 194.5, 



GiV6n i&=233."5 7 ; 

r A = 39° 47' 36", 
to find \ B = 50° 12' 24", 

[ c = 303.9. 

Construct triangle ABC, having the given 
parts. 

f tan ^4 = a/6, 
Formulas \ B = 90° - A, 

[ c = 6 sec ^4 = 6/cos^l. 




Fig. 32 



ISOSCELES TRIANGLES 71 

log a = 12.28892 - 10 log b = 12.36829 - 10 

log b = 2.36829 log cos A = 9.88557 - 10 

.-. log tan A = 9.92063 - 10 .\ log c = 2.48272 

... A = 39° 47' 36". .-. c = 303.89. 
.-. B = 50° 12' 24". 

Observe that the subtraction above is simplified by writing the char- 
acteristic 2 of log a and log b in the form 12 — 10, and the characteristic 
— 1 of log cos A in the form 9 — 10. 

Check. b 2 = (c + a) (c - a). log (c + a) = 2.69758 

Here c + a = 498.4, log (c - a) = 2.03902 

c- o = 109.4. .-.log 6 = 4.73660/2 

= 2.36830. 

As this computed value of log b differs by only .00001 from that found 
in the table, the computed parts are probably correct to four places. 



EXERCISE XVII 

Solve the triangle ABC, having given : 

1. 5 = 67°, a = 5. 9. a = 3.414, 6 = 2.875. 

2. A= 38°, a = 8.09. 10. A = 46° 23 r , c = 5278.6. 

3. J. = 15°, c = 7. 11. a = 529.3, c = 902.7. 

4. B = 50°, 6 = 20. 12. 5 = 23° 9 7 , b = 75.48. 

5. a = 0.35, C = 0.62. 13. B = 18° 38 x , c = 2.5432. 

6. a = 273, 6 = 418. 14. ^1 = 31° 45 r , a = 48.04. 

7. b = 58.6, c = 76.3. 15. 6 = 617.57, c = 729.59. 

8. A = 9°, 6 = 937. 16. 5 = 82° 6 r 18", a = 89.32. 

50. Isosceles triangles. In an isosceles triangle the perpen- 
dicular from the vertex to the base divides the isosceles triangle 
into two equal right triangles. Hence any two parts which 
determine one of these right triangles determine also the 
isosceles triangle. 



72 



PLANE TRIGONOMETRY 



In this and the next article we shall use the following nota- 
tion in isosceles triangles : 

r = one of the equal sides, 

c = base, 

h = altitude, 

A = one of the equal angles, 

C = angle at the vertex, 

Q = area of the triangle. 

Given r and c; to find A, C, h, 

ISO -2 A, 




B " Ex. 1. 

Fig. 33 and Q. 

A = cos- 1 (c/2 r), C: 



Also, 



h = Vr 2 - (c/2) 2 = V(r + c/2)(r- c/2). 
/i = r sin J. or ^ = (c/2) tan J.. 

Q = cA/2. 



51. Regular polygons. Lines drawn from the center of a 
regular polygon of n sides to the vertices divide the polygon 
into n equal isosceles triangles ; and the perpendiculars from 
the center to the sides of the polygon divide these n equal 
isosceles triangles into 2n equal right triangles. 

Hence any two parts which determine one of these equal 
right triangles determine also the regular polygon. 

Using the notation given in fig. 
34, we have 

C/2 = 360° /(2n) = 180° /n. 

If p = the perimeter of the poly- 
gon and F = the area, we have 

p = nc, F =ph/2. 

CA, or r, is the radius of the cir- 
cumscribed circle and CD, or A, is 
the radius of the inscribed circle. 




REGULAR POLYGONS 73 



EXERCISE XVIII 

In an isosceles triangle, having given : 

1. c and A ; find C, r, ft. 

2. ft and C ; find A, r, c. 

3. c and ft; find A, C, r. 

4. c = 2.352, C = 69° 49' ; find r, ft, A, Q. 

5. ft = 7.4847, J. = 76° 14 / ; find r, c, C, Q. 

6. A barn is 40 x 80 ft. , the pitch of the roof is 45° ; find the length 
of the rafters and the area of both sides of the roof, the horizontal 
projection of the cornice being 1 ft. 

7. One side of a regular decagon is 1 ; find r, ft, F. 

8. The perimeter of a regular dodecagon is 70 ; find r, h. F. 

9. In a regular octagon ft = 1 ; find r, c, F. 

10. The area of a regular heptagon is 7 ; find r, ft, p. 

11. The side of a regular octagon is 24 ft. ; find ft and r ; also find the 
difference between the areas of the octagon and the inscribed circle, and 
the difference between the areas of the octagon and the circumscribed 
circle. 

12. The side of a regular heptagon is 14 ft. ; find the magnitudes as 
in example 11. 

13. Each side of a regular polygon of n sides is c ; show that the radius 
of the circumscribed circle is equal to (c/2) esc (180° /?i), and the radius 
of the inscribed circle is equal to (c/2) cot (180°/^) • 

14. The radius of a circle is k ; show that each side of a regular 
inscribed polygon of n sides is 2 k sin (180°/ n), and that each side of a 
regular circumscribed polygon is 2 k tan (180°/ n). 

15. The area of a regular polygon of sixteen sides inscribed in a circle 
is 100 sq. in. ; find the area of a regular polygon of fifteen sides inscribed 
in the same circle. 

16. The radius of a circle is 10 ; find the area between the perimeters 
of two regular polygons of thirty -six sides each, one circumscribing the 
circle and the other inscribed in it. 



CHAPTEE V 



SOLUTION OF TRIANGLES IN GENERAL 



52. The two following relations which the sides of any 
triangle bear to the sines and cosines of its angles are funda- 
mental in the study and solution of triangles. They are 
called the law of sines and the law of cosines respectively. 

Law of sines. The sides of a triangle are proportional to the 
sines of their opposite angles. 




Let A, B, C denote the numerical measures of the angles of 
the triangle ABC, and a, b, c the numerical measures of its 
sides. 

From C draw CD A- BA, or BA produced. 

In fig. x, A is acute; in fig. y, A is obtuse. 

In each figure we have 

by §7, sinB=p/a, (1) 

by §21, sin A = DC/ AC =p/b. (2) 

Dividing the'members of (2) by those of (1), we obtain 

sin A /sin B = a/b, or a / sin A = b /sin B. (3) 

Similarly, or by symmetry from (3), we obtain 

a/sin A = c/sin C, or 6/sin B = c/sin C. 
14 



LAW OF COSINES To 

Hence -A-r = -i- = -^-. [23] 

sin A sin B sin C L J 

Observe that if C = 90°, sin C = 1 and [23] gives 
a/c = sin A and &/c = sin B, 
which are the known relations in the right-angled triangle. 

In § 62 each ratio in [23] will be shown to be equal to the 
diameter of the circle circumscribed about the triangle ABC. 

Law of cosines. In any triangle the square of any side is 
equal to the sum of the squares of the other two sides minus 
twice the product of these two sides into the cosine of their 
included angle. 

In figures 35 regard AD, DB, and AB &s directed lines. 
Then in each figure we have 

AD + DB = AB) .". DB = c - AD. (1) 

Squaring both members of (1) and adding ^> 2 , we obtain 

(DB 2 + p 2 ) =c 2 + (AD 2 +p 2 ) -2c- AD. (2) 
In each figure DB 2 + p 2 = a 2 , AD 2 +/> 2 = b 2 , 
and AD j AC = cos A. 

Whence AD = b cos .4. 

Substituting these values in (2), we obtain 

a 2 = b 2 + c 2 - 2 be cos A. [24] 

Similarly, or by symmetry from [24], we obtain 

b 2 = a + c 2 — 2 ac cos B, (1) 

c 2 = a + b 2 -2ab cos C. (2) 

Observe that if A = 90°, [24] becomes a 2 = b 2 + c 2 , which 
is the known relation between the sides when .4 = 90°. 
Solving [24] for cos A, (1) for cos B, etc., we obtain 

b 2 + c 2 - a 2 „ a 2 + c 2 - b 2 ro ^_, 

cos A = ■ , cos B = , etc. |2o | 

2 be 2ac L J 



76 PLANE TRIGONOMETRY 

The form of the law of cosines in [25] is useful in finding 
any angle of a triangle from its sides. 

53. If of the six parts of any triangle we have given any 
three (one, at least, being a side), the triangle, as we know by 
Geometry, is determined and can be constructed. Hence, in 
the numerical solution of triangles by Trigonometry, we must 
consider the following four cases, the given parts being : 

(i) One side and two angles. 

(ii) Two sides and the angle opposite one of them. 

(iii) Two sides and their included angle. 

(iv) Three sides. 

In solving triangles we frequently use the two following 
geometric properties of triangles : 

I. The sum of the three angles is equal to 180°. 
II. The greater angle is opposite the greater side, and vice 
versa. 

54. Cases (i) and (ii). If two of the three known parts of 
a triangle are a side and its opposite angle, a fourth part can 
evidently be found by the law of sines. 

Hence cases (i) and (ii) can be solved by the properties I 
and II in § 53 and the law of sines. 



(A =65°, 
Ex. 1. Given \ B = 40°, to find 



r C = 75°, 
\ 6 = 35.46, 
a = 50; [ c = 53.29. 



Construct the triangle BAG having the 
given parts. 

f C = 180° - (A + B) = 75°, 
Formulas \ b = asmB/siuA, (1) 

^ € = a sin C/sin A. (2) 

Using the table in § 5, we obtain from (1) 
Fig. 36 and (2), 




OBLIQUE-ANGLED TRIANGLES 



77 



6 = 50 sin 40° /sin 65° 
= 50 x 0.6428/0.9063 = 35.46, 
and c = 50 x 0. 9659 /.0. 9063 = 53.29. 

Check. By construction and measurement. 
Ex. 2. Given A = 50°, C = 65°, c = 30 ; find B, a, 6. 

r^!=60 o , f£ = 45°, 

Ex. 3. Given i a = 3 y/2, to find iC= 75°, 
[ 6 = 2 V3 [ c = 4.73. 

Construct the triangle ABC, having the 
given parts. 

Observe that only one such triangle can 
be constructed. 



fsin B = b sin A /a, (1) 

Formulas \ C = 180° - (A + B), 



c = a sin (7 /sin -4 
Since b < a, 5 < ^4, ie. £ < 60°. 
From (1), 
Hence, by § 10, 




(2) 



Fig. 37 



V2 



sin B = ^^ sin 60° = 

3 V2 2 



5 = 45°, since B < 60°. 

C = 180° - (A + J?) = 180° - 105° = 75°. 
From (2), c = 3V2x 0.9659 -f- ( V3/2) = 4.73. 

Check. By construction and measurement. 
Ex. 4. If C = 60°, a = 2, c = V6, find 6, 4, 5. 
Ex. 5. If A = 30°, a = 9, b = 6, find J5, O, c, having given 
sin 19° 28' = 1/3 and sin 130° 32' = 0.76. 

55. Cases (iii) and (iv). If two sides of a triangle and their 
included angle are known, the third side can be found by [24] ; 
if the three sides are known, each angle can be found by [25]. 
Hence cases (iii) and (iv) can be solved by the law of cosines. 



A = 60°, 



[ a = 7, 



Ex. 1. Given 



6 = 8, to find < B = cos- 1 (1/7), 
c = 5; I C = cos- 1 (11/ 14). 



78 PLANE TRIGONOMETRY 

( a 2 = b 2 + c 2 — 2 6c cos J., (1) 

Formulas \ cos B = (a 2 4- c 2 - 6 2 ) / (2 ac), (2) 

[cos C = (a 2 + & 2 - c 2 ) / (2 a&). (3) 

From (1), a 2 = 8 2 + 5 2 - 2 - 8 • 5 - (1/2) = 49. 

From (2), cos B = (7 2 + 5 2 - 8 2 )/(2 ■ 7 • 5) = 1/7. 

From (3), cos C = (7 2 + 8 2 - 5 2 )/(2 . 7 - 8) = 11/14. 

r.a = 7, B = cos- 1 (1/7), C = cos- 1 (11/ 14). 

Check. By construction and measurement. 

Ex. 2. If a = 7, 6 = 3, c = 5, find A, £, 0, having given 
11/14 = cos 38° 13', 13/14 = cos 21° 47'. 

1)2 + C 2 _ a 2 3 2 + 52 _ 72 ! 

cos J. = = = 

2 be 2-3-5 2 

T , a 2 + c 2 - 6 2 7 2 + 5 2 - 3 2 13 

cos B = = = — 

2 ac 2-7-5 14 

a 2 + 6 2 - c 2 7 2 + 3 2 - 5 2 11 

cos = = = — • 

2 ab 2-7-3 14 

.-. A = 120°, 5 = 21° 47', C = 38° 13'. 
Check. A + B+ C = 120° + 21° 47 / + 38° 13' = 180°. 

Ex. 3. If a = 2, 6 = 3, c = 4, find ^L, J5, C, having given 

7/8 = cos 28° 57', 11/16 = cos 46° 34', 1/4 = cos 75° 31'. 

56. Since the law of cosines involves sums, it is not adapted 
to computation by logarithms. Hence, to solve cases (iii) and 
(iv) by logarithms, we must deduce other formulas, one of 
which is the law of tangents below. 

Law of tangents. The sum of any two sides of a triangle is 
to their difference as the tangent of half the sum of their oppo- 
site angles is to the tangent of half their difference. 

From the law of sines, we have 

a/b = sin A /sin B. (1) 



LAW OF TANGENTS 79 

By principles of proportion, from (1), we obtain 

a + b _ sin A -f- sin B 
a — b sin A — sin B 

_ 2sinH^+*)cosH^-*) by r 191 r2Q1 
" 2 cos \ (A + B) sin J (,i - J5) by Li JJ ' LZUJ 

= tan i (A + B) /tan J (A - B). [26] 

Since tan \ (A + £) = tan i (180° - C) 

= tan (90° - C /2) = cot (C/2), 
from [26] we obtain 

A — B a — b C _„- 

tan = -cot-. [2<] 

2 a + b 2 L J 

As a check, [26] is the more convenient form, while for solving 
triangles, [27] is the preferable form of this law. 

Example. If a = V3, b = 1, C = 30°, find A, B, c, having given 
cot 15°= (V3+ 1)/(V3- 1). 

By [27], tan £_z_^ = ^ cot C 

J L J 2 a + 6 2 

= V3 " 1 cotl5°=l. 

V3 + 1 

Hence i(^-^)= 45 °- (!) 

Also, | (^L + B) = \ (180° - C) = 75°. (2) 

Adding (1) and (2), A = 120°. 

Subtracting (1) from (2), B = 30°. 
Since C = .B, we have c = 6 = 1. 

Check. By construction and measurement. 

SOLUTION OF OBLIQUE TRIANGLES WITH LOGARITHMS 

57. Case (i). Given one side and two angles. 

f a = 180, | ( C = 66° 17', 

Ex. 1. Given + A = 38°, to find << b = 283.33, 

I 5 = 75° 43' : ! c = 267.68. 



80 



PLANE TRIGONOMETRY 



Construct triangle ABC, having the given parts. 



Formulas 



Logarithmic formulas 



f C = 180° - (A + B) = 66° 17', 
b = a sin B/sinA, 



<j b = a 
I c = a 



sin O/sin A. 



( log b = log a + log sin B — log sin .4, 
1 log c = log a + log sin C — log sin A. 




Fig. 38 



loga= 2.25527 
log sin J5 = 9.98636 - 10 
12.24163 - 10 
log sin A = 9.78934 -10 
/.log b= 2.45229 

.-. b = 283.33 



b + a _ tan \ (B + A) 



Check. 



log (6 + a) = 2.t 

\og(b-a) = 2.01423 

log quotient = .65166 



6 — a tan J (J5 — J.) 

6 + a = 463.33. 
6 - a = 103.33. 



loga= 2.25527 
log sin C = 9.96168 - 10 
12.21695- 10 
log sin A = 9.78934 - 10 
.-. logc= 2.42761 

.-. c = 267.68 



(.B-f^)/2 = 56 o 51'30". 

(5-^4)/2 = 18°51 / 30 // . 

log tan \(B + A) = 10.18514 - 10 

log tan \{B-A)= 9.53347 - 10 

log quotient = .65167 



(1) 



As the logarithms of the two members of (1) differ by only 1 in the 
fifth place, the value of b is correct to four places. 
Similarly we can check the value of c. 



C = 54°30 / , 


a = 100. 


C = 62° 5', 


b = 100. 


C = 78° 10', 


a = 102. 


5 = 65°, 


c = 270. 


B = 29° 17', 


C = 135°. 


^1 = 44°, 


C = 70°. 



OBLIQUE TRIANGLES 81 

EXERCISE XIX 

Solve each of the following triangles : 

1. Given B = 60° 15', 

2. Given A = 45° 41', 

3. Given B = 70° 30', 

4. Given A = 55°, 

5. Given a = 123, 

6. Given b = 1006.62, 

7. A ship S can be seen from each of two points A and B on the 
shore. By measurement, AB = 800 ft., Z SAB = 67° 43', and Z SBA 
= 74° 21' 16". Find the distance of the ship from A. 

8. A flag pole A is observed from two points B and C, 1863 ft. 
apart. Given Z BCA = 36° 43 r and Z CBA = 57° 2V, find the distance 
of the flag pole from the nearer point. 

9. To determine the distance of a hostile fort A from a place B, a 
line BC and the angles ABC and BCA were measured and found to be 
322.55 yd., 60° 34', and 56° 10' respectively. Find the distance AB. 

10. A balloon is directly over a straight level road, and between two 
points on the road from which it is observed. The points are 15847 ft. 
apart, and the angles of elevation are found to be 49° 12 r and 53° 29' 
respectively. Find the distance of the balloon from each point of 
observation. 

11. To find the distance from a point itoa point B on the opposite 
side of a river, a line AC and the angles CAB and ACB were measured 
and found to be 315.32 ft., 58° 43", and 57° 13' respectively. Find the 
distance AB. 

12. From points A and B, at the bow and stern of a ship respectively, 
the foremast, C, of another ship is observed. The points A and B are 
300 ft. apart; the angles ABC and BAC are found to be 65° 31" and 
110° W respectively. What is the distance between the points A and C 
of the two ships ? 



82 



PLANE TRIGONOMETRY 



58. Case (ii). Given two sides and an angle opposite one of 

them. Let a, b, A be the given parts. Then, to find B, C, c, 

we have • «».,,, 

sin B = b sin A /a, (1) 

C = 180° -(A + B), 

c = a sin C /sin A. 

Since two supplementary angles have the same sine (§ 30), 
the relation in (1) gives in general two values for B, both of 
which are to be taken unless one is excluded by the conditions 
of the problem. 

We have to consider the three following cases : 

I, when a > b ; II, when a = b\ III, when a < b. 

I. When a > b and A is acute or obtuse, then A > B ; hence 
B is acute, and there is but one triangle having the given parts, 

II. When a = b and A is acute, then A = B ; hence B is acute 
and the triangle is isosceles. 

In this case the triangle can be solved by the method in § 50 or by the 
law of sines and the relation A + B + C — 180°. 

If a = b and A = or > 90°, the triangle is impossible. Why ? 

Note. Example 1 below and the first three examples in Exercise XX 
may be solved before Case III is considered. 




B 2 ~~][)" By 



^ A 



y 


O 


b/ 


a 


A m 


B^ 



D 



^X A 




Fig. 39 



III. When a < b and A is acute, there are two triangles, one, 
or no triangle, having the given parts, according as a >, =, or 
< b sin A. 

Geometric proof. In each figure, let Z. XA C = A and AC = b. 



OBLIQUE TRIANGLES 83 

Draw CD lil; then in each figure CD = b sin A. 

With C as a center and a as a radius, describe the arc mn. 

If a > b sin A (i.e. if a > CD), the arc mrz, will cut AX 
(fig. as) in two points, B x and i? 2 , on the side of A toward D, 
and there will be two unequal triangles having the parts a, b, 
A, viz., the triangles ACB X and ACB 2 . Hence B has the two 
values Z. AB X C and Z. AB 2 C, which are supplementary. 

If a = b sin A, the arc mn will touch AX at D (fig. y)\ 
hence B = 90° and only the right-angled triangle .4 CD has 
the given parts. 

If a < b sin A, the arc mn will not meet .4X (fig. z)\ hence 
no triangle can be constructed with the given parts. 

E.g., if a = 5, b = 7, and ^4 = 30°, then a < b and a > b sin ^4 ; hence 
there will be two triangles having these parts. 

If a < b and A = or > 90°, the triangle is impossible. Why ? 

Trigonometric proof. From the law of sines, 

sin B = b sin A /a, or (b/a) sin A. (1) 

Also .4 < B and .4 + B < 180°. (2) 

If a > b sin A, b sin A/a<l, whence from (1), sin B < 1 ; 
hence B has two unequal values, which are supplementary. 

Since a<b, b/a > 4, whence from (4), sin B > sin .4 ; hence 
each of the two supplementary values of B will satisfy both 
the conditions in (2). 

Therefore B has two values and there are two different 
triangles having the given parts (fig. x). 

If a = b sin A, sin B = 1 in (4), whence B = 90° ; hence the 
required triangle is right angled at B (fig. y). 

If a < b sin A, sin B > 4, which is impossible; hence the 
triangle is impossible (fig. z). 

From the trigonometric proof, it follows that if a < b and A 
is acute, there are two triangles, one, or no triangle, according 
as log sin B is negative, zero, or positive. Why ? 



84 



PLANE TRIGONOMETRY 




Ex. 1. Given 



to find 



f a = 250, 
J 6 = 240, 

[A = 72° 4'; 

f JB = 65° 58' 24" 
J. C = 41° 67' 36" 
[ c = 175.69. 



Here a>b and 4 < 9.0° ; hence B < 90° and there 
B is only one triangle having the given parts. 
Fig. 40 Construct the triangle ABC, having the given 

parts. 

f sin B = b sin A / a, 
Formulas i C = 180° - (A + B), 
I c = a sin C/sin^L. 



log 6= 2.38021 
log sin A = 9.97837 - 10 
12.35858 - 10 

log a = 2.39794 
log sin B = 9.96064 - 10 

.-. B = 65° 58 / 24". 

C = 180° 



loga = 2.39794 
log sin C = 9.82517 - 10 
12.22311 - 10 
log sin A = 9.97837 - 10 

logc 



Check. 



b + c 



2.24474 

c = 175.69. 
(72° 4' + 65° 58' 24") = 41° 57' 36' 

toni(B+C) 



b-c tani(JB-CT) 

Here 6 + c = 415.69, 

&-c»64.31, 

log (b + c) = 2.61877 

log (6 - c) = 1.80828 

log quotient = .81049 



(1) 



l(B+ C) = 53°58 / ; 
i(5- C)= 12°0'24". 

log tan i (B + O) = 10. 13821 - 10 

log tan i (B - C)= 9.32772 - 10 

log quotient = .81049 



As the logarithms of the two members of (1) are equal, the values 
obtained above are correct. 

Ex, 2. How many triangles are there which have the following parts ? 

(i) a = 70, b = 90, A = 30°. 

(ii) a = 40, b = 80, 4 = 30°. 

(iii) a = 20, b = 50, J. = 30°. 

(iv) a = 70, b = 75, 4 = 60°. 



OBLIQUE TRIANGLES 



85 



f a = 732, rJ5 = 63° 2' 20" or 116° 57' 40", 

Ex.3. Given ^ 6=1015, to find J C = 76° 57' 40" or 23° 2 / 20 // , 
I ^. = 40° ; I c = 1109.4 or 445.66. 

Here a < b and a > 6 sin A ; hence by III there are two solutions. 
Construct the two triangles ACB\ and ACB 2 , having the given parts. 

Formula for B. sin B = b sin A /a. 

log 6= 3.00647 
log sin A = 9.80807 - 10 



12.81454 
loga= 2.86451 



10 



.-.log sin B= 9.95003-10 

.-. B t = 63° 2' 20", B 2 = 116° 57' 40". 

To find the unknown parts of ACB\, we have 

ZACB X = 180° - (A + Bi) = 76° 57' 40". 
^4-Bi = a sin A CBi/sin A. 

JO 




Fig. 41 



To find the unknown parts of ACB 2 , we have 

Z ACB 2 = 180° -(A + B 2 ) = 23° 2' 20" 
^12?2 =. a sin ^4 CJ5 2 / sin J.. 



loga= 2.86451 
log sin A CB l = 9.98866 ■ 



12.85317 - 10 
9.80807 - 10 



Check. 



log sin A 
.-AogAB 1 = 3.04510~~ 
.-. AB X = 1109.4. 

c + fr _ tan|(C + £) 
c - 6 _ tan|(C- J5)" 



loga= 2.86451 
JL0 log sin A CB 2 = 9.59257 - 10 
12.45708 - 10 
log sin ^4 = 9.80807 - 10 



log AB 2 = 2.64901 
.-. AB 2 = 445.66. 



(1) 



86 PLANE TRIGONOMETRY 

Here c + 6 = 2124.4, J (C + jB) = 70° ; 

c-b= 94.4, i (C - B) = 6° 57' 40". 

log (c + b) = 3.32723 log tan \ (C + 5) = 10.43893 - 10 

log(c - 6) = 1.97497 logtani(C - B) = 9.08670 - 10 
log quotient = 1.35226 log quotient = 1.35223 

The equality of these logarithms to five figures verifies the answers to 
four figures. 

EXERCISE XX 

Solve the following triangles : 



1. 


Given a = 145, 


b = 178, 


JB = 


41° 10'. 


2. 


Given b = 573, 


c = 394, 


B = 


112° 4'. 


3. 


Given a = 5.98, 


6 = 3.59, 


A = 


63° 50'. 


4. 


Given a = 140.5, 


b = 170.6, 


^4 = 


40°. 


5. 


Given 6 = 74.1, 


c = 64.2, 


C = 


27° 18'. 


6. 


Given a = 27.89, 


6 = 22.71, 


B = 


65° 38'. 


7. 


Given b = 45.21, 


c = 50.3, 


B = 


40° 32' 7". 


8. 


Given a = 34, 


6 = 22, 


B = 


30° 20'. 


9. 


Given a = 55. 55, 


6 = 66.66, 


B = 


77° 44' 40' 


0. 


Given a = 309, 


6 = 360, 


A = 


21° 14' 25' 


.1. 


Given b = 19, 


c = 18, 


C = 


15° 49'. 



59. Case (iii). Given two sides and their included angle. 

Let a, b, C be the given parts. Then, to find A, B, c, we have 

A-B a-b C _ 

tan __ = __ cot _, (1) 

(,4+£)/2 = 90°- C/2, (2) 

c = a sin C/sin A. (3) 

From (1) we obtain (A — B)/2. Having given (A — B) / 2 
and (A + B)/2 9 we readily obtain A and B. 
Then c can be found by (3). 

(a = 540, r^i = 78°17 / 40", 

Example. Given J 6 = 420, to find -J B = 49° 36' 20", 

[c=52°6 / ; [ c = 435.15. 



OBLIQUE TRIANGLES 87 

f A-B a-b C 

tan = cot — i 

Formulas 2 a + 6 2 

I (.4 + £)/2 = 90°- C/2 = 63° 57', 

L c = a sin C/sinA. 

Here a + 6 = 960, a - b = 120, C/2 = 26° 3'. 

log (a -6)= 2.07918 loga = 2.73239 

log cot (C/2) = 0.31086 log sin C = 9.89712 - 10 

12.39004 - 10 12.62951 - 10 

log (a + b) = 2.98227 log sin A = 9.99087 - 10 

.-. log tan i(A- B)= 9.40777 - 10 ,\ log c = 2.63864 

.-. (A -B)/2 = 14° 20' 40". .-. c = 435.15. 

Also (A + £)/2 = 63°57'. 

.-. A = 78° 17' 40", 

and B = 49° 36' 20". 

Check. A + 5 -f C = 52° 6' + 78° 17' 40" 4- 49° 36' 20" = 180°. 

To check c we could use c = b sin C/sin 5. 

EXERCISE XXI 

Solve each of the following triangles : 

1. Given a = 266, b = 352, C = 73°. 

2. Given 6 =91.7, c = 31.2, A = 33° 7' 9". 

3. Give* a = 960, 6 = 720, C = 25° 40'. 

4. Given a = 886, b = 747, C = 71° 54'. 

5. Given 6 = 41.02, c = 45.49, ^ = 62° 9' 38". 

6. Two trees A and B are on opposite sides of a pond. The dis- 
tance of A from a point C is 297.6 ft., the distance of B from C is 
864.4 ft., the angle AC Bis 87° 43' 12". Find the distance AB. 

7. Two mountains A and B are respectively 9 and 13 miles from a 
town C, and the angle ACB is 71° 36' 37". Find the distance between 
the mountains. 

8. Two points A and B are visible from a third point C, but not 
from each other. The distances AC, BC, and the angle ACB were 
measured and found to be 1321 ft., 1287 ft., and 61° 22' respectively. 
Find the distance AB. 

9. From a point 3 mi. from one end of an island and 7 mi. from the 
other end, the island subtends an angle of 33° 55' 15". Find the length 
of the island. 



88 



PLANE TRIGONOMETRY 



10. Two stations A and B on opposite sides of a mountain are both 
visible from a third station C. The distance AC, BC, and the angle 
ACB were measured and found to be 11.5 mi., 9.4 mi., and 59° 31' 
respectively. Find the distance from A to B. 

11. Two trains leave the same station at the same time on straight 
tracks that form an angle of 21° 12'. Their average speeds are 40 mi. 
and 50 mi. an hour respectively. How far apart will they be at the end 
of the first thirty minutes ? 

60. Case (iv). Given the three sides. To solve this case we 
first obtain formulas for the sine, cosine, and tangent of A/ 2, 
B/2, and C/2 in terms of the three sides a, b, c. 

Let 2s = a + b + c. 

Then 2 (s — a) = b + e — a, 

2 (s — b) = a + e — b, 

2 (s — e) = a + b — c. 

By [16] and [17], we have 

sin 2 (^/2) = (l-cos^)/2, 
COS 2 (yl/2)EE(l + cos^)/2. 

Substituting for cos A its value given in [25], we obtain 



(i) 



sin 2 7r = ^ 1- 



: + c 2 



2* 

a 2 - (b 



2 be 



COS^ 






b* + c 2 - a 1 



2 be 

n 2 



4 be 

(a — b -\- c)(a + b — c) 
4 be 

4:(s-b)(s- e) 



4 be 

(b + c + a) ( b + c — a) 



sin 



■ t = v 

Dividing 



4 be 

(s-b)(s-c) 



be 



• . cos 



4: be 
_ 4 s (s — a) 
4 be 

s(s — a) 



tan 



-4 

A _ / (s - b) (s - c) 

2 -\ s (s - a) 



be 



by(i) 



k28] 



THREE SIDES GIVEN 89 

Observe that s is the half-perimeter, a the side opposite, and 6 and c 
the sides including the angle A. 

Ex. 1. State in words the relations in [28]. 

Ex. 2. Write the values of the sine, cosine, and tangent of B/2; (7/2. 

Since the trigonometric ratios of A/ 2, B/2, C /2 must all 
be positive (?), only the positive roots in [28] are taken. 

Formulas [28] enable us to obtain any angle of a triangle 
from the sides, either through its sine, its cosine, or its tangent. 

Ex. 3. If a = 13, 6 = 14, c = 15, find A, B, C, having given 
1/2 = tan 26° 34', 4/7 = tan 29° 45', 2/3 = tan 33° 41'. 

Here s = (13 + 14 + 15) /2 = 21, s - a = 21 - 13 = 8, 

s- 6 = 21 - 14 = 7, s - c = 6. 

Hence tan — = x /— = - = tan 26° 34'. 

2 \21-8 2 



26° 3 

B 18-6 4 

Again, tan — = a / = - = tan 29° 4o . 

6 ' 2 \21-7 7 



.-. .4/2 = 26° 34', or A 



.-. B/2 = 29° 45', or 5 = 59° 30'. 
Similarly C/2 = 33° 41', or C = 67° 22'. 

CTecA:. 4 + B + C = 53° 8' + 59° 30' + 67° 22' = 180°. 

Ex.4. If a = 35, 6 = 84, c = 91, find A, B, C, having given 
tan 11° 19' = 1/5, and tan 33° 41' = 2/3. 

Ex. 5. If a = 13, 6 = 14, c = 15, find the sines of A/2, B/2. C/2. 

If we multiply the value of tan (4/ 2) in [28] by 1 in the 
form Vs — a I "V ' s — a, the expression for tan (.1/2) can be 
put in the more symmetrical form 

. l(i - a )( s - b )( s - c ) 



■- = — ^ 

2 s - a \ 



Letting r = x /(s - a) (s - b) (s - c)/s, [29] 

we have tan (A/ 2) = r/(s — a).T 

Similarly tan (B/2) = r/ (s - b), I [30] 

tan (C/2) = r/(s - c).J 



90 PLANE TRIGONOMETRY 

When all the angles are required, the tangent formulas [30] 
are the best, as they involve the fewest logarithms. 

In § 63 it will be proved that r in [29] is the radius of the 
circle inscribed in the triangle ABC. 

f a = 130, f A = 77° 19' 9", 

Ex. 6. Given \ b = 123, to find J 5 = 67° 22' 49", 

[ c = 77 ; [C = 35° 18' 2". 



Formulas 



r = V(,s — a) (s — b) (s — c) /s, 
tan (-4/2) = r/(s - a), 
< tan(B/2) =r/(s-6), 
tan(C/2) =r/(s - c). 



Here s = 165, s - a = 35, s - 6 = 42, s - c = 88. 

log (s-a) = 1.54407 .-. log tan (A/2) = 9.90309 - 10, 

log (s - b) = 1.62325 whence ^4/2 = 38° 39 7 34.6" ; 

log (s -c) = 1.94448 log tan (B/2) = 9.82391 - 10, 

5.11180 whence 5/2 = 33° 4V 24.4"; 

log s = 2.21748 log tan ( C / 2) = 9. 50268 - 10, 

.-. log r = 2.89432/2 whence (7/2 = 17° 39' 1". 

= 11.44716 - 10. 

Check. A + B + C = 77° 19' 9" + 67° 22' 49" + 35° 18' 2" = 180°. 

EXERCISE XXII 

Solve each of the following triangles : 

1. Given a = 56, b = 43, c = 49. 

2. Given a = 8.5, b = 9.2, c = 7.8. 

3. Given a = 61. 3, 6 = 84.7, c = 47.6. 

4. Given a = 705, b = 562, c = 639. 

5. Given a = .0291, 6 = .0184, c = .0358. 

6. Given a = 85, & = 127, .4 = 26° 26'. 

7. Given a = 5.953, b = 9.639, C = 134°. 

8. Given a = 3019, b = 6731, c = 4228. 

9. Given a = 60.935, c = 76.097, A = 133° 41'. 



THREE SIDES GIVEN 91 

10. Given b = 74.8067, c = 98.738, C = 81° 47'. 

11. Given b = 129.21, c = 28.63, A = 27° 13'. 

12. Given a = 2.51, 6 = 2.79, c = 2.33. 

13. Given a = 32.163, c = 27.083, C = 52° 24' 16". 

14. Given a = 74.8, c = 124.09, 5 = 83° 26' 52". 

15. Given a = 86.0619, c = 63.5761, A = 19° 12 / 43". 

16. Given a = 93.272, 6 = 81.512, C = 58°. 

17. The sides of a triangular field are 534 ft., 679.47 ft., and 474.5 ft. 
Find the angles. 

18. A pole 13 ft. long is placed 6 ft. from the base of an embankment, 
and reaches 8 ft. up its face. Find the slope of the embankment. 

19. A point P is 13581 in. from one end of a wall 12342 in. long, 
and 10025 in. from the other end. What angle does the wall subtend at 
the point P ? 

20. The distances between three cities A, B, and C are as follows : 
AB = 165 mi., AC = 72 mi., and BC = 185 mi. B is due east from A. 
In what direction is C from A? 

21. Under what visual angle is an object 7 ft. long seen when the eye 
of the observer is 5 ft. from one end of the object and 8 ft. from the 
other end ? 



22. Prove sin A = 2 Vs (s - a) (s - b) (s - c) / (be). 

23. Prove cos A — s (s — a) / (be) — (s — b) (s — c) / (be) . 

24. If a = 18, b = 24, c = 30, show that sin A = 3/5. 

25. The sides of a triangle can be substituted for the sines of their 
opposite angles, and vice versa, when they are invoked homogeneously in 
the numerator and denominator of a fraction, or in both members of an 
equation. 

For this substitution is the multiplication of each term by the equal 
numbers, a /sin A, b /sin B, c/sin C, or their reciprocals. 

E.g.. sin A = sin (180° - A) - sin (B + C). 

.-. sin A = sin B cos C + sin C cos B. by [7] 

Multiplying the first term by a /sin A, the second by 6/sin B. and the 
third by c/sin C, we obtain 



92 PLANE TRIGONOMETRY 

a = b cos C + c cos B. 1 
Similarly b = a cos C + c cos -4, > (1) 

and c = a cos i? + b cos ^i. J 

rt „ _ 2 c 2 + a 2 2 sin 2 C + sin 2 J. 

26. Prove = 

3 abc 3 b sin A sin 

27. Multiplying the equations in (1) of example 25 by — a, 6, and c 
respectively, and adding, we obtain the law of cosines, 

6 2 + c 2 - a 2 = 2bccosA. 

28. Prove the relations in (1) of example 25 directly from a figure. 
Suggestion. See the figures in § 52. 

61. Area of a triangle. 

(i) In terms of two sides and the sine of their included angle. 
Let F denote the area of any triangle, as A BC in § 52. 
Then F =BA • DC/2 = be sin A/2. [31] 

(ii) In terms of the three sides. 
By [31] and [13], we have 

F = be sin (.4/2) cos (A /2). 
Hence, by [28], F = Vs(s - a) (s - b) (s - c). [32] 

(iii) In terms of one side and the sines of the angles. 
In [31], putting for c its value b sin C /sin B (obtained from 
the law of sines), we have 

F= b'sinAsinC 

2 sin B L J 

EXERCISE XXIH 

1. State [31], [32], [33] in words. 

2. Given A = 75°, b = 10, c = 40 ; to find F. 

3. Find the areas of the triangles in examples 1-4 in Exercise XXI. 

4. Find the areas of the triangles in examples 1-4 in Exercise XX. 

5. Find the areas of the triangles in examples 1-4 in Exercise XXII. 

6. Find the areas of the triangles in examples 1-4 in Exercise XIX. 



THREE IMPORTANT CIRCLES 



93 




62. Circumscribed circle. Let D denote the diameter of the 
circle circumscribed about the triangle 

ABC. 

Through C draw the diameter COH, 
and join HB. 

Then £A=Z.H, 

Z CBH = 90°, 
and CH=D. 

.'. sin A = sin H = a/D. 
.'.a/smA = D. (1) 

Compare (1) with [23]. 

Example. Find the radii of the circles circumscribed about the 
triangles in examples 1-4 in Exercise XIX. 

63. Inscribed circle. Let r denote the radius of the circle 
inscribed in the triangle ABC. Join the center with the 
points of contact D, E, F. Draw OA, OB, OC. 

C 



Fig. 42 




Then OD = OE = OF = r. 

A ABC = A BOC + A COA + A AOB. 
.'. F = ar/2 + br/2 + cr/2 
= r (a + b + c) J '2 = rs. 



§60 



94 



PLANE TRIGONOMETRY 



Hence r = F/s, 

or by [32], r = V(s - a) (s - b) (s - c)/s. (1) 

Compare (1) with [29]. 

Example. Find the radii of the inscribed circles of the triangles in 
examples 1-4 in Exercise XXII. 

64. An escribed circle of a triangle is a circle which is 
tangent to one side of the triangle and to each of the other 
two sides produced. 

Ex. 1. If r a denotes the radius of the escribed circle tangent to the 
side a, prove that 

r a = F/(s - a)= Vs (s — b)(s - c) / (s — a). 
A ABC = AACD + A ABB - A BCD. 
.:F = r a (b + c-a)/2 
= r a (s -a). 




Fig. 44 



Ex. 2. Find the radii of the three escribed circles of the triangles in 
examples 1-3 in Exercise XXII. 



CHAPTER VI 

RADIAN MEASURE, GENERAL VALUES, TRIGONOMETRIC 
EQUATIONS, INVERSE FUNCTIONS 

65. A radian is an angle which, when placed at the center 
of a circle, intercepts between its sides an arc equal in length 
to the radius of the circle. 

That is, if arc AB is equal to the D ^- -.^ / p . 

radius OA, then 

angle AOB = a radian. 




66. Constant value of the radian. 
In fig. 45, let 

arc AB = OA = r units. 
Then, by Geometry, we have the following proportion : 
A AOB: 180° = arc .473 : semicircumference, 
i.e. a radian : 180° = r : irr. 

From this proportion, we have 

tt radians = 180° = 2 right angles. [34] 

Formula [34] expresses the relation between radians, degrees, 
and right angles, and should be fixed in mind. 

From [34], a radian =(180/7r)°, or (180/3.1.416)° (1) 

= 57°.295779, or 57°.3, approximately (2) 

= 57° 17' 44".8 ? approximately. (3) 

From [34], 1° = (tt/180) radian, (4) 

or a right angle = 90° = (tt/2) radians. (5) 

Hence 270° = (3 tt/2) radians, 

and 360° = 2 ir radians. (6) 

95 



96 PLANE TRIGONOMETRY 

Ex. 1. Express in radians the angle 8° 15'. 

8° 15' = (33/4)° = (33/4) (ar/180) radian by (4) 

= 0. 144 radian, approximately. 

Ex. 2. Express in radians the sum of the three angles of a triangle. 
Ex. 3. Express in degrees the angle 5/8 radian. 

5/8 radian = (5/8) (57° 17' 44". 8) by (3) 

= 35° 48' 35". 5. 

67. Radian measure of angles. In fig. 45, by Geometry, we 
have 

Z AOC : ZAOB = arc ABC : arc AB. (1) 

Let Z A OC = N radians, 

arc ABC — s units, 
and arc AB = OA = r units. 

Substituting these values in proportion (1), we obtain 
N radians : 1 radian = sir. 

.\N = s/r. [35] 

That is, the number of radians in an angle at the center of a 
circle is equal to the intercepted arc divided by the radius. 

Cor. If r = l, N = s. 

The number of radians in an angle is called its radian (or 
circular) measure. 

When the measure of an angle is given in terms of it or any 
other numeral or numerals and no angular unit is expressed, 
the radian is always understood as the unit. 

E.g., the angle 2 is an angle of 2 radians, and the angle 
7r/2 is an angle of 7r/2 radians. 

The fraction 22/7 gives the correct value of it to two places of deci- 
mals, and for many purposes this value is sufficiently accurate. 

Ex. 1. An angle at the center of a circle whose radius is 5 ft. inter- 
cepts an arc of 3 ft. Find the angle in degrees. 

The angle = (3/5) radian § 67 

= (3/5) (180/tt) = 34° T 4 T . § 66 



RADIANS 97 

Ex. 2. An angle of 42° 20' is at the center of a circle whose radius is 
10 ft. Find the length of the arc intercepted between its sides. 

Let x = the number of feet in the intercepted arc. 

Then a/10 = the number of radians in 42° 20', or (127/3)° 

= (127/3) (*/ 180). by [34] 

.-. x = (127/54) (22/7) = 1 T \%, approximately. 
Hence the intercepted arc is approximately 7 ± \\ ft. long. 

Ex. 3. From [35] find s in terms of N and r, and r in terms of s 
and N. 

EXERCISE XXIV 

1. Express in radians two positive and two negative angles each of 
which is coterminal with 7T/4; 5 7T/4; 3tt/2; 5?r/2; 7t/S; 2?r/3; 
?r/6; 5?r/6. 

Express in degrees each of the following angles : 

2. 2tt/3. 3. 5;r/3. 4. 5 it. 5. 3tt/4. 

Express in radians each of the following angles : 

6. 45°. 8. 90°. 10. 97° 25'. 12. 43° 25' 36". 

7. 135°. 9. 270°. 11. 175° 13'. 13. 38° 17' 23". 

Find the complement and the supplement of : 

14. 7T/4. 15. 2tt/3. 16. 3?r/4. 17. 5tt/3. 

Find the trigonometric ratios of each of the following angles : 

18. 7T/6. 19. Tt/4:. 20. 7t/3. 21. tt/2. 22. it. 

23. Two angles of a triangle are 1/2 and 2/5. Find the third angle 
in radians, also in degrees. 

24. The difference between the two acute angles of a right-angled 
triangle is n/1. Find the angles in radians, also in degrees. 

25. Express in radians, also in degrees, the angle subtended at the 
center of a circle by an arc whose length is 15 ft. , the radius of the circle 
being 20 ft. 

26. The diameter of a graduated circle is 6 ft., and the graduations 
on its rim are 5' apart. Find the distance from one graduation to the 
next. 



98 PLANE TRIGONOMETRY 

27. Find the radius of a globe which is such that the distance between 
two places on the same meridian whose latitudes differ by 1° 10' may be 
half an inch. 

28. The value of the divisions on the outer rim of a graduated circle 
is 5' and the distance between successive graduations is 0.1 in. Find 
the radius of the circle. 

29. Assuming the earth to be a sphere and the distance between two 
points 1° apart to be 69J mi. , find the radius of the earth. 

68. Principal values. If sin = — 1/2, we know that is 
any angle which is coterminal with — 7r/6 or — 57r/6. Of 
this series of values the smallest in size, — tt/6, is called the 
principal value of 0. 

The principal value of an angle having a given trigonometric 
ratio is the angle, smallest in size, which has this ratio. 

Hence, if sin or esc is positive, the principal value of 
lies between and 7r/2; if sin or esc is negative, the 
principal value of lies between — 7r/2 and 0. 

If tan or cot is +, the principal value of lies between 
and 7r/2; if tan or cot is — , the principal value of 
lies between — 7r/2 and 0. 

If cos or sec is +, the principal value of lies between 
and 7r/2, preference being given to the positive value; if 
cos or sec is — , the principal value of lies between 7r/2 
and ir. 

69. Angles having the same sine. 

Let Z XOP = A, Z XOP f = w - A. 

Then sin XOP 1 = sin XOP, esc XOP 1 = esc XOP, 

and any angle which has the same sine or cosecant as A is, 
P' p or can be made, coterminal with A 

x. yS or 7r — A. 

Ns^-^i! jr I n this chapter n will denote 

~m* ^^ ' ■ J— x any positive or negative integer, 

Fig. 46 including zero. 



TRIGONOMETRIC EQUATIONS 99 

When n is even, nir + A includes all the angles, and only 
those, which are coterminal with A. 
Now, when n is even, 

nir + A = nir + (- 1)M. (1) 

Again, when n is odd, n — 1 is even, and (n — 1) ir + (ir — A) 
includes all the angles, and only those, which are coterminal 
with ir — A. 

But when n is odd, 

(n - 1) 7T + (tt - A) =mr — A = mr + (- l) n A. (2) 

From (1) and (2) it follows that the expression nir + (— 1)M 
includes all the angles, and only those, which are coterminal 
with A or ir — A. 

Hence the general expression nTT+(— l) n A includes all the 
angles, and only those, which have the same sine or cosecant 
as A. 

That is, the general value of 8 in the equation 

sin 6 = sin A (or esc 6 = esc ^4) is nir + (— l) n A. 

E.g., if sin = sin(7T/3), then 6 = nit + (- l) n 7t/S. 

Example. Find the general value of 0, if sin 6 = — 1/2. (1) 

The principal value of 6 in (1) is — it / 6. 

Substituting for — 1/2 in (1) its identical expression sin(— tt/6), we 
obtain the equivalent equation 

sin0 = sin(- 7t/6). (2) 

.-. e = n7t + (-l) n (-it/6) 

= mt-(- 1)«- 7t/6. (3) 

By using the principal value of 6 in (2), we obtain the simplest expres- 
sion for 6 in (3). 

Sometimes, however, in solving an equation it is advantageous to use 
some other than the principal value of the unknown angle. 

Observe that the unit understood with 6 and A is the radian. If the 
degree were the unit, we would write B = n • 180° + (— l) n A. 

LofC. 



100 



PLANE TRIGONOMETRY 




Fig. 47 



70. Angles having the same cosine. 

IP Let Z XOP = A, Z XOP 1 = - A. 

Then cos XOP 1 = cos XOP, 

^ and sec XOP 1 = sec XOP, 

and any angle which has the same cosine or 
Jp' secant as A is, or can be made, coterminal 
with A or — A. 
Now 2n7r ± A includes all the angles, and only those, 
which are coterminal with A or — A. 

Hence 2 rnr ± A includes all the angles, and only those, 
which have the same cosine or secant as A. 

That is, the general value of 8 in the equation 

cos = cos A (or sec 6 = sec A) is 2 nir ± A. 

E.g., if cos 6 = cos (tt/9), then $ = 2 nit ± tt/9, or 2 n • 180° ± 20°. 
Example. Find the general value of 0, if cos = — V3/2. 
Substituting for — V3/2 its identical expression cos (5 zr/6), we obtain 
the equivalent equation 

cos0 = cos (5 it / 6). 
.-. = 2 ntf ± 5 7T/6, or 2 n • 180° ± 150°. 

71. Angles having the same tangent. 

Let Z XOP = A,£ XOP 1 = tt + A. 

Then tan XOP 1 = tan XOP, cot ZOP' = cot XOP, 

and any angle which has the same tangent or cotangent as 
A is, or can be made, coterminal with 
A or it + A. 

When n is even, nir + A includes M / y\ \— \ x 

all the angles, and only those, which 
are coterminal with A. 

When n is odd, n — 1 is even, and 
(n — 1) 7r + (tt + -4), i.e. ?i7r + A, includes all the angles, and 
only those, which are coterminal with it + A. 









P 


M' 










JSo 


M ' 


jP' 


Fig. 48 







TRIGONOMETRIC EQUATIONS 101 

Hence nir + A includes all the angles, and only those, which 
have the same tangent or cotangent as A. 

That is, the general value of 8 in the equation 

tan = tan A (or cot = cot A) is nir + A. 

E.g., if cot = cot(— 7T/5), = mt + (- #/5), or mt — 7t/5. 

Example. Find the general value of 0, if tan = — y/Z. 
Substituting for — V3 its identical expression tan(— tt/3), we obtain 
the equivalent equation 

tan — tan(— tt/3). 
.-. d = n7t - it/%, or n • 180° - 60°. 

72. A trigonometric equation is an equation which involves 
one or more trigonometric ratios of one or more unknown 
angles, as tan = 1 or 2 sin 2 + ^/3 cos + 1 = 0. 

To solve a trigonometric equation is to obtain an expression 
for all the angles which will satisfy it. 

The first step in solving a trigonometric equation is to solve 
it algebraically for some trigonometric ratio of the unknown 
angle ; the second step is to apply one or more of the princi- 
ples in §§ 69-71. 

Some elementary types of trigonometric equations are solved 
below. 

Ex. 1. Solve the equation sin 2 = 1/4. 
First step. sin = ± 1 /2. 

Second step. .-. sin = sin (± 7t/6). 

.-. e = n7t + (- 1)»(± tt/6) = mt ±it/6. §69 

Ex. 2. Solve the equation cos 2 = 1/2. 

Denote the two values of cos by cos X and cos 2 . 

Then cos B x = V2 / 2 = cos (it / 4) , (1) 

and cos 2 = - V2 /2 = cos (5 it /4). (2) 

From (1), 0i = 2 ?ii?r ± tt/4. (3) 

From (2), 2 = 2 n 2 it ± 5 tt/4 = (2 n 2 ±l)it ± ar/4. (4) 



102 PLANE TRIGONOMETRY 

When n is even, nit ± ?r/4 includes 2 ni7t ± 7T/4 in (3) ; and when n 
is odd, nit ± 7T/4 includes (2 n 2 ± 1) it ± it / '4 in (4). 

Hence = mt ± ?r/4. 

Observe that, by using 5 ?r/4 instead of the principal value of 2 in (2), 
the two expressions for in (3) and (4) are more readily combined into 
one expression. 

Ex. 3. Solve the equation 2 sin 2 + V3 cos + 1 = 0. (1) 

Putting for sin 2 its identical expression 1 — cos 2 0, we obtain the 
equivalent equation 

2 cos 2 - V3 cos 6 - 3 = 0, (2) 

which involves only one function of the unknown angle 0. 

Factor, (cos - V3) (2 cos + V3) = 0. (3) 

By Algebra, (3) is equivalent to the two equations 

cos = V3, cos = - V3/2. 

Now, by § 26, cos = V3 is impossible, 

and cos = - V3/2 = cos(5 7T/6) 

gives = 2n7T ± 5 7zr/6. §70 

Ex. 4. Solve the equation tan 5 = cot 2 0. 

Substituting for cot 2 its identical expression tan(?r/2 — 2 0), we 
obtain the equivalent equation 

tan 5 = tan(?r/2 - 20). 
.-. 50 = n7t -\-(7t/2 -2 0). §71 

.'. = (7l7t + 7T/2)/7. 

Ex. 5. Solve tan sec = - V2. (1) 

Putting for sec0 its identical expression Vtan 2 + 1, we obtain the 
equivalent equation 

tan Vtan 2 0+1 = - y/2. (2) 

Square, tan 4 + tan 2 = 2. 

Factor, (tan 2 + 2) (tan 2 - 1) = 0. 

Hence tan = ± V^2 or ± 1. (3) 

Now tan = + V— 2 is impossible. 

Since tan sec is negative in (1), tan and sec are opposite in 
quality, whence is in the third or fourth quadrant. 

Hence tan = + 1 gives = 2 7i7r + 5#/4, 

and tan = — 1 gives = 2 mt — it /4. 



TRIGONOMETRIC EQUATIONS 103 

Observe that the solutions of (3) which do not satisfy (1) were intro- 
duced by squaring (2). 

Ex. 6. Given sin = — 1/2 and tan — 1/ V3 ; to find the general 
value of 0. 

Since sin is — , and tan is +, must be in the third quadrant. 

The smallest positive angle in the third quadrant which will satisfy 
sin = — 1/2 is 7 7T/6, and this angle satisfies also tan = 1 / V3. 

Hence = 2 mt + 7 ?t/6, or 2 n • 180° + 210°. 

EXERCISE XXV 

Solve each of the following equations : 

1. sin 2 = 1. 3. tan 2 0=1. 5. cos 2 = 1/4. 

2. esc 2 = 2. 4. cot 2 = 3. 6. sec 2 = 4/3. 

7. 2sin 2 + 3 cos = 0. 10. sin 2 - 2 cos + 1/4 = 0. 

8. cos 2 - sin = 1 /4. 11. sin -f cos = s/2. 

9. 2 V3 cos 2 = sin 0. 12. 4 sec 2 0-7 tan 2 = 3. 

13. tan + cot = 2. 

14. tan 2 - (1 + V3) tan + V3 = 0. 

15. cot 2 + (V3 + 1/V3)cot0 + 1 =0. 

16. tan 2 + cot 2 = 2. 

17. tan + sec = 3. 22. sin 2 = cos 3 0. 

18. 2 sin = 1 -f cos 0. 23. cos m = sin r 0. 

19. sin50 = l/V2. 24. sin cos = 1/2. 

20. cos 50 = cos 4 0. 25. sin 0cos = - V3/4. 

21. cot = tan r 0. 26. sec esc = - 2. 

27. tan 2 tan = 1. 

Find the most general value of that satisfies : 

28. cos = - 1 / V2 and tan = 1. 

29. cot = - V3 and esc = - 2. 

30. If cos(A- B) = l/2 and sin (A + B) = l/2, find the smallest 
positive values of A and B and also their general values. 



104 PLANE TKIGONOMETRY 

31. If tan (.4 - JB) = 1 and sec ( A + B) = 2 / V3, find the smallest 
positive values of A and B and also their general values. 

73. The addition and subtraction formulas and those for 
the sum and difference of sines or cosines are often useful in 
solving certain types of trigonometric equations. 

E.g., take the equation a sin 9 + b cos 6 — c. (1) 

Let A denote the principal value of tan -1 (a/b) ; then tan A = a/b, 
and from the fundamental relations in § 24 we obtain 



sin^i = a/Va 2 + & 2 , cos .4 = 6/ Va 2 + b 2 . (2) 

Dividing both members of (1) by Va 2 + b 2 and substituting for the 
resulting coefficients of sin and cos their values given in (2), we have 



sin A sin + cos A cos = c / V«2 _j_ &2 # (3) 



Let B d enote the principal value of cos - 1 (c/'wa 2 + b 2 ) ; then 
cos B = c/Va 2 + 6 2 , and by [10] from (3) we obtain 

cos (0 — A) = cos B. 

.-. - A=2n7t ±B. §70 

.-. = 2mt + A ± B, or 2 n • 180 + J. ± B. (4) 

In (4) we have the solution of any equation of the type (1) when a 
and b are real and c < or = Va 2 + 6 2 , arithmetically. 

Ex. 1. Solve 4 sin + 3 cos = 5, given tan 53° 7' 45" = 4/3. 
Here A = tan- 1 (4/3) = 53° 7' 45", 

and 5 = cos- 1 (5 / V4 2 + 3 2 ) = cos- 1 1 = 0. 

.-. = 2ft,.18O o +53°7 / 45". 

Ex. 2. Solve sin + sin 5 = sin 3 0. (1) 

By [19], sin0 + sin 50 = 2 sin 3 cos 20. (2) 

From (1) by (2), 2 sin 3 cos 2 = sin 3 0. 

.-. sin 3 0(2 cos 20 - 1) = 0. 
From sin 3 = 0, 3 = nit. 

From cos 2 = 1/2, 2 = 2 ntf ± 7t/S. 

Hence = 7i7T/3 or mt ± 7t/6. 



TRIGONOMETRIC EQUATIONS 105 

EXERCISE XXVI 

Solve each of the following equations : 

1. sin + sin 7 = sin 4 0. 7. sin + sin 3 + sin 5 = 0. 

2. cos + cos 3 = cos 2 0. 8. cos + cos 2 + cos 3 = 0. 

3. sin0 - sin 3 = cos 2 0. 9. V3 cos + sin = V2. 

4. cos + cos 7 = cos 4 0. 10. sin + cos = V2. 

5. sin 4 - sin 2 = cos 3 0. 11. v^sin - cos = V2. 

6. sin 7 = sin + sin 3 0. 12. sin + cos0 = y/2 cos(7r/5). 

13. 5 sin + 2 cos = V29, given tan 68° 12' = 5/2. 

14. 2 sin - 3 cos = V13/2, given tan 33° 41' 24" = 2/3. 

74. Trigonometric functions. A quantity whose value depends 
upon one or more other quantities is called a function of these 
quantities. 

E.g., the circumference or the area of a circle is a function of the 
radius ; the area of a rectangle is a function of the base and the altitude ; 
the volume of a rectangular parallelopiped is a function of the three 
dimensions. 

Since the trigonometric ratios of an angle depend upon the 
size of the angle, they are often called the trigonometric func- 
tions of the angle. 

Beside the six trigonometric functions defined in § 21, there 
are two others which are frequently used : 

1 — cos A is called the versed sine of A, written vers A. 
1 — sin A is called the coversed sine of A, written covers A. 
As a ratio, vers A = (OP — OM) / OP, 

and covers A = (OP - MP) / OP. 

Note. After the analogy of vers A and covers A many other trigo- 
nometric ratios, or functions, of A might be invented and named. 

E.g., 1 + cos A, or (OP + OM) / OP, is sometimes called the suversed 
sine of A, and sec A — 1, or (OP — OM) / OM, the external secant of A. 



106 PLANE TRIGONOMETRY 

Observe that the trigonometric lines defined in § 32 are also 
functions of the angle or of its measuring arc. 

But by the trigonometric functions we usually mean the 
trigonometric ratios. 

75. Inverse trigonometric functions. 

By § 23, if c = sin 0, then 6 = sin" 1 * (1) 

As the value of sin 6 depends on the value of the angle 0, 
so the value of sin _1 c depends on the value of c. 

Hence, on the one hand, the sine is a function of the angle, 
and, on the other hand, the angle is a function of the sine. 

An angle is often called the inverse function of any one of 
its trigonometric ratios. Thus sin _1 c is often read inverse 
sine c ; cos -1 c, inverse cosine c ; tan _1 c, inverse tangent c, etc. 

Note. If A denotes all the angles which have the same sine, then 
sin— 1 (sin A) = A and sin (sin- 1 c) = c ; hence, if we regard sin as denot- 
ing the operation of finding the sine of and sin -1 as denoting the operation 
of finding the angle whose sine is, then sin~ l and sin denote inverse 
operations, i.e. operations each of which undoes what the other does. 
For this reason sin— l c is called the inverse sine of c. 

Sin— l c is read also anti-sine c, or arc siyiec. 

For each value of 0, sin has a single definite value. 

For each value of c, sin -1 c has, by § 23, an indefinite number 
of values. 

Thus the trigonometric functions are single-valued, while the 
inverse trigonometric functions are multiple-valued. 

E.g., if tan = 1, = tan— 1 1 = nit -f tt/4, and the principal value of 
tan— l l is 7T/4. 

If cos = 1/2, 6 = cos— l (1 /2) = 2 n7T ± 7T/3, and the principal value 
of cos- 1 (l/2) is 7T/3. 

If sin0 = V2/2, 6 = sin- 1 (V2/2) = nit + (- 1) w tt/4, and the prin- 
cipal value of sin- 1 (V 2 /2) is tt/4. 

Example. If a and c are positive and c < 1, and the inverse functions 
are restricted to their principal values, show that 

sin— x c + cos - l c — it /2, tan— l a + cot -1 a = tt/2. 



INVERSE FUNCTIONS 107 

76. Sum and difference of two inverse tangents. 

To prove tan -1 m ± tan -1 n = tan -1 • [361 

1 1 zp mn L J 

Let A = tan -1 m, B = tan -1 ^. (1) 

Then tan A = m, tan B = n. (2) 

Using the notation in (1), [36] becomes 

. m ± w / . , ^s m ztn 

A ±B = tan" 1 > or tan (.4 ± £) = ; (3) 

1 zp mn 1 if: mw x 7 

To prove (3), by § 37 we have 

tan A ± tan B 



tan (4 ± B) 



1 ip tan A tan i? 
= (m ± n) / (1 zp mn). by (2) 



The proof above illustrates the method of procedure in 
establishing relations between inverse functions. 

Ex. 1. Prove that sin- 1 (3/5) + sin- 1 (8/17) = sin- 1(77/ 85). (1) 

Let A = sin- 1 (3/6), B = sin- 1 (8/17). (2) 

Then sin A = 3/5, sin B = 8/17. (3) 

.-. cos ^4 = 4/5, cos B =15/17. (4) 

Using the notation in (2), (1) becomes 

A + B = sin- 1 (77/85), or sin (A + B) = 77/85. (5) 
To prove (5), by [7] we have 

sin (A + B) = sin A cos B + cos A sin J5 

4MM- **« 

Ex. 2. Prove that cos- 1 (4/5) + tan- 1 (3/ 5) = tan- 1 (27/11). (1) 

Let A = cos- 1 (4/ 5), 5 = tan- 1 (3/5). (2) 

Then cos A = 4/5, tanJ. = 3/4, tanJB = 3/5. (3) 
Using the notation in (2), (1) becomes 

A + 5 = tan- 1 (27/ 11), or tan (A + B) = 27/11. (4) 



108 PLANE TRIGONOMETRY 



To prove (4), by [11] and (3) we have 

* i a , t* 3/4 + 3/5 27 

tan (A + B) = = — • 

V ' 1- (3/4) (3/5) 11 

Ex. 3. Prove that 2 tan- 1 (1/3) + tan- 1 (1/7) = ?r/4. (1) 

Let -4 — tan- 1 (1/3), B = tan~ 1 (l/7). (2) 

Then tanA = l/3, tan B =1/7. (3) 

Using the notation in (2), (1) becomes 

2^ + E=7r/4, or tan(2 4 + JB) = tan(*r/4) = l. (4) 

x> rim + o „ 2 tan J. 2/3 3 ■ 

By [151, tan 2 JL = = = - • (5) 

J L J 1 -tan 2 A 1-1/9 4 w 

To prove (4), by [11], (3), and (5) we have 

tan (2,1 + 3) ee 8/4 + 1/7 =1. 
V ' 1- (3/4) (1/7) 

Ex. 4. Solve the equation tan- 1 2 x + tan- 1 3 x = 7T/4. (1) 

By §76, tan- 1 2 x + tan- 1 3 x = tan- 1 2 a + 8 ^ ■ (2) 

From (1) by (2), tan~ l 2x + 8x = - . (3) 

W J V ; ' l-6x 2 4 W 

T7 /ox 2x + 3x , it 

From (3), = tan - = 1. 

1 ■ 1 -6x 2 4 

.\x = 1/6 or - 1. 

x = 1/6 satisfies (1) for the principal values of tan— l 2x and tan— x 3x. 

x = — 1 satisfies (1) for the values 

tan- 1 (-2) = 116°33 / 55 // , 

tan- 1 (-3) = -71°33 / 55 // . 

££ _J_ "^ Ofc \ 

Ex. 5. Solve the equation tan- 1 h tan~ x = tan -1 (—7). (1) 

x — 1 x 

.05+1 "■ .x-1 . .2x 2 -x + l ■ 

tan- 1 h tan- 1 = tan- l (2) 

x — 1 x 1 — x 

From (1) by (2), 2x2 ~ x + 1 _ _ 7? r x = 2. 
1 — x 

x = 2 satisfies (1) when for tan- 1 (- 7) we take 98° 7' 48". 



INVERSE FUNCTIONS 



109 



EXERCISE XXVII 

1. Find the value of vers (it / 6) , vers (it / 4) , vers (it / 3) , vers (3 it / 4), 
versO, vers (it / 2), vers rt, vers (3 it/ 2). 

2. Find the value of covers (?r/6), covers (#/ 4), covers (#/3), 
covers (3 7r/ 4), covers 0, covers (it/ 2), covers 7T, covers (3 7t/2). 

3. Express in radians the general value of 

sin- 1 (V2/2); sin- 1 (- V3/2) ; cos~ 1 (V3/2); cos- 1 (-1/2); 
tan-^VS/S); tan-^-V^); cot~ 1 (-l); cot-^v^/S). 

Prove each of the following relations for principal values : 
L 2 + tan- 1 0.5 = it/ 2. 

7 + tan-13 = 153° 26' 6".5, given 0.5 = tan 26° 33' 53".5. 
, ?n — n _ 7T 
= 4* 



4. tan - 

5. tan- 

6. tan~ 



13. tan- 

14. tan— 

15. tan— 

16. tan- 

17. cos- 



tan - 



n m -f n 

7. tan- 1 (1/7) + tan- 1 (1/13) = tan- 1 (2/9). 

8. 2 tan- 1 (2/3) = tan- 1 (2/3) + tan- 1 (2/3) = tan- 1 (12/5). 

9. tan- 1 (3/4) + tan- 1 (3/5) - tan- 1 (8/ 19) = it/4. 

Add tan- 1 (3/ 5) to tan- 1 (3/ 4) and then subtract tan- 1 (8/19). 

10. sin- 1 (3/5) + sin- 1 (8/17) = sin- 1 (77/85). 

11. cos- 1 (4/5) + cos- 1 (12/13) = cos- 1 (33/65). 

12. cos- 1 (4/ 5) + tan- 1 (3/ 5) = tan- 1 (27/ 11). 
Solve each of the following equations : 

x + tan- 1 (1 - x) = tan- 1 (4/3). 



+ tan~ 



x- 2 x + 2 4 

x + 2cot~ 1 x = 2 7T/3. 

(x + 1) + tan- 1 (x - 1) = tan- 1 (8/31). 



-f tan~ 



2x 



2it 
3 ' 



x 2 + 1 " x 2 - 1 

18. sin-ix + sin- 1 2x = it/S. 

19. sin- 1 (5/x) + sin- 1 (12/x) = it/2. 

Suggestion. Observe that sin- 1 (5/x) and sin -1 (12 /x) are comple- 
mentary angles. 



CHAPTEE VII 



PERIODS, GRAPHS, IMPORTANT LIMITS, COMPUTATION OF 
TABLE, HYPERBOLIC FUNCTIONS 

77. Periods of the trigonometric functions. As an angle 
increases from to 2 it radians, its sine first increases from 
to 1, then decreases from 1 to — 1, and finally increases 
from — 1 to 0. As the angle increases from 2 it to 4 it radians, 
the sine again goes through this same series of changes. Thus 
the sine goes through all its changes while the angle increases 
by 2 it radians. This is expressed by saying that the period 
of the sine is 2 ir. 

Similarly the cosine, secant, or cosecant goes through all 
its changes while the angle increases by 2 it radians. 

The tangent or cotangent, however, goes through all its 
changes while the angle increases from to it radians. 

Hence the period of the sine, cosine, secant, or cosecant is 
2 ir radians; while the period of the tangent or T 

cotangent is IT radians. 

Since, as the angle increases, each trigonometric 
function goes through again and again the same 
series of values, these functions are called periodic 
functions. 



78. Curve of sines. 

and 

Then, if 
we have, when 

0=0, tt/6, tt/3, 
sin 0=0, M X P X , M 2 P 2 , OP z 



Let OX= OPi = + 1, 
Z.XOP 1 = tt/6, 
ZXOP 2 = *tt/3. 

6=Z XOP, 




h 



it/2, 2 tt/3, 5tt/6, it, 7 tt/6, 4 tt/3, 
-M 1 P 1 , ~M 2 P 2 



M 2 T 2 , M 1 P 1 , 0, 
110 



CURVE OF SINES 



111 



The series of values through which sin 6 passes as the angle 6 
increases can be graphically represented by means of the points 
of a curve constructed as follows : 

Let OX and OY in fig. 50 be two fixed straight lines at right 
angles to each other, and let an angle of one radian be repre- 
sented by a unit of length along OX. Take OR = it units, 
i.e. about 3\ units of length ; also take RX = tt units, and 
OX' = — 7r units. 

Then OR represents it radians ; OX, 2 tt radians ; and OX', 
— it radians. 









Y 


p ,I >" P <n 












P 


^T^n s p I¥ 






-Tt 


- 2 ^3 -\ -% 


d/\ 


\X"\ \^7T 


*% 3 % 5 % 


2 7r 


x'\ 


! i | i 


\/\ N 


\ \ 2?P 3 7 ^\ 


J ' ! 


/x 




^^J^ \^T 




r' 




^V^ 1 i ^f 





Fig. 50 



Lay off on OX the distances representing 0, 7r/6, 7r/3, tt/2, 
2 tt/3, etc. At N, the end of tt/6, draw NP ± OX and equal 
to M 1 P 1 in fig. 49; at the end of tt/3 draw a perpendicular 
equal to M 2 P 2 ; and so on. Through the ends P, P', P", • • • of 
these perpendiculars draw a smooth curve. 

Then if from any point in this curve, as P, we draw PN _L OX, 
the directed line NP represents the sine of the angle whose 
radian measure is represented by the directed line OX. 

This curve, called the curve of sines, or sinusoid, consists 
of portions similar to OP"RQX placed one after another. 

This illustrates graphically that the period of the sine is 
2tt. 



79. Curve of cosines. LTsing fig. 49, Ave obtain, when 
6=0, tt/6, tt/3, it 12, 2 tt/3, 5 tt/6, tt, 7 tt/6, 4 tt/3, - 
cos<9= + l, OM u OM 2 , 0, -OM 2 , -OM x ,-l, -OM v -OM 2r 



112 



PLANE TRIGONOMETRY 



Lay off on OX the distances representing 0, tt/6, 7r/3, 7r/2, 
2 7r/3, etc. At the points thus determined erect perpendiculars 
equal to OX, OM v OM 2 , 0, — OM 2 , etc., and trace a curve through 
the points P, P\ P", etc. 

The curve thus obtained is called the curve of cosines. 







P 










"l ' 


X' 


-%/ 


i i 


i | \/2 /3 


7T 


4 ^3 


y i 


i jz 




i y 
i y 


■%.-% 


r' 


1 
i 


1 1 


' 3 ^ 2 


27T 



Fig. 51 



Observe that the curve of cosines has the same form as the 
curve of sines. These two curves differ only in their position 
with reference to the origin 0. 

80. Curve of tangents. Using fig. 49, where XOP = 2ir /5, 
we obtain, when 

(9=0, tt/6, tt/3, 2tt/5, tt/2, -tt/6, -tt/3, -2tt/5, -tt/2,. ■ ., 
tan (9=0, ITi, 1^ XT, oo, -ZZ^ -1T 2j -XT, -oo, 

Proceeding as in § 78, we obtain the curve represented by the 
continuous lines in fig. 52. This graph is called the curve of 
tangents. 

Since tan (it / 2) = oo, the curve of tangents will have no 
point on RK, but the infinite branch P'P" will approach 
indefinitely near to the line RK without ever touching it. 
The same is true of the infinite branch P 2 P 3 with reference 
to the line R-^K^ 

The curve of tangents will evidently consist of an unlimited 
number of similar but disconnected branches, any one of 
which is parallel to the branch P 3 OP ft . 



CURVE OF TAXGENTS 



113 



Curve of cotangents. Using fig. 49, by § 30, we obtain, when 

(9=0, tt/10, tt/6, tt/3, tt/2, 2tt/3, 5tt/6, 9tt/10, it, •••, 
cot0=oo, XT, XT 2 , XT l9 0, ~XT 1} -XT 2 , -XT, -oo, • ••. 



."!# 




Fig. 52 



Proceeding as above, we obtain the curve represented by 
the dotted lines in fig. 52 as three of the infinite number of 
disconnected branches of the curve of cotangents. 

81. Curve of secants. Using fig. 49, we obtain, when 

0=0, tt/6, tt/3, 277-/5, tt/2, -tt/6, -77-/3, -2tt/5, -77-/2,- ■ •, 
secO= + l,OT 1 , OT 2 , OT, oo, OT u OT 2 , OT, oo, 

Proceeding as above, we obtain BPK and B'P'K' as two 
branches of the curve of secants; repetitions of these two 
branches make up the rest of the curve. 



114 



PLANE TRIGONOMETRY 



Similarly the curve of cosecants can be constructed, three 
branches of which are represented by the dotted curved lines 
in fig. 53. 



27T 




Fig. 53 

82. *Lt(sin 9/8) = lt(tan8/9) = 1, when 8 = 0, the unit of 
6 being the radian. 

Let be the radian measure of any positive acute angle 
A OP. Draw the arc AP, the chord AP, PM _L OA, AT Jl OA. 

* When, according to its law of change, a variable will become and remain 
less in size than any assignable constant value but will never become zero, the 
variable is called an infinitesimal, or is said to approach zero as its limit. 

When a variable approaches a constant so that their difference is an infini- 
tesimal, the variable is said to approach the constant as its limit, or the 
constant is said to be the limit of the variable. 

Lt (sin e /e) is read the limit of sin 0/0; = is read approaches as its 
limit or is an infinitesimal. 

The reciprocal of an infinitesimal is an infinite and is denoted by oo. 

Any number which is neither an infinitesimal nor an infinite is called a, finite 
number 




LIMIT OF SIS 0/6 115 

Then A OAP < sector OAP < A OA T. (1) 

By Geometry, the area of the sector OAP is OA -arc AP/2. 
Hence, from (1), we have 

OA ■ MP < OA • arc AP < OA • A T. (2) 

Dividing each member of (2) by OA 2 , we obtain 
MP J OP < arcAP/OA < AT/OA ; 
that is, sin < < tan 0. (3) 

Dividing the members of (3) first by sin and then by tan 0, 
we have 

1< 0/sin < sec (9, (1) P, 

and cos < 5 /tan < 1. 

Let = 0. 

Then sec 9 = 1, 

and cos = 1. 

Hence the ratio 0/sin or the ratio 0/tan0 lies between 
1 and a variable whose limit is 1 ; hence the limit of each of 
these ratios is 1. 

Now - 0/sin (-0) =- 0/(- sin 0) = 0/sin 0. 

Hence It [- 0/sin (- 0)J = lt(0/sin0) == 1, when = 0. 
Also, lt[- 0/tan(- 0)] = It (0/tan 0) = 1. 

Hence the theorem holds true whether is positive or 
negative. 

Example. To find sin V and cos V. 

By § 66, V = (0.00 029 088 821 + ) radian. (V) 

By (3) and (1'), sin V < V in radians < 0.00 029 088 822. (2 7 ) 

Again, cos V = Vl-sin 2 !/ > Vl - (.0003) 2 > .99 999 99. 

Hence, to seven places, 

cos r = 0.99 999 99 + . (3 7 ) 

By (4), sin d > cos 6. 

.-. sin V > (r in radians) cos 1'. 



116 PLANE TRIGONOMETRY 

.-. sin V > 0.00 029 088 821 x 0.99 999 99 by (1'), (30 

> 0.00 029 088 821 x (1 - 0.00 000 01), 
or sin r> 0.00 029 088 820 + . (4') 

From (2') and (4') it follows that to ten places 

sin V = 0. 00 029 088 82 + , (50 

the eleventh figure being or 1. 

This example affords a good illustration of the use of corollary 1, below. 

Two very important corollaries to the theorem above are the 
following : 

Cor. 1. If 6 is very small, it can be substituted for either 
sin 8 or tan 8 in approximate calculations ; or vice versa. 

Cor. 2. If 8 is an infinitesimal and it is substituted for either 
sin 8 or tan 8 in any function, the limit of the function will not 
be changed. 

83. Convenient formulas for computing the sine and cosine of 
any angle are 

d> 3 <f> 5 d> 7 d> 9 d> n 

sm ^-| + |-*+|_| ?+ .., (b) 

<f> 2 d> 4 <f> 6 <f> 8 <f> 10 
cos^l-| + |-| + |-^ + .., (c) 

where <f> is the radian measure of the angle. 

For the proof of (b) and (c), see § 94 in Taylor's Calculus. 

The sine of any angle is arithmetically equal to the sine or 
cosine of some angle not greater than 7r/4 ; hence its value can 
be computed by taking <£ in (b) or (c) not greater than tt/4. 

Likewise the cosine of any angle can be computed by taking 
<f> in (c) or (b) not greater than 7r/4. 

Observe that for <£ < 7r/4 the series in (b) or (c) converges 
rapidly, and only a few terms need to be computed. 

Having found the value of the sine and the cosine, the log- 
arithmic sine and cosine can be obtained from a table of 
logarithms of numbers. The logarithmic tangent can then be 



SIMPSON'S METHOD 117 

found by subtracting log cos from log sin, and log cot by 
subtracting log sin from log cos, or log tan from 0. 

Example. Compute sin 11° 12' 23" and cos 11° 12' 23". 
By §66, 

11° = (0.01 745 329 252 x 11) radian = 0.19 198 621 772 radian. 
12' = (0.00 029 088 821 x 12) " = 0.00 349 065 852 " 
23" = (0.00 000 484 814 x 23) " =0.00 011150 722 " 
.-. = 11° 12' 23" =0.19 558 838 346 " 

Substituting this value for in (b) and (c), from the first three terms 
we obtain 

= 0.19 558 838 1 = 1.00 000 000 

0V[5 = .00 000 238 071 4= .00 006 098 

.19 559 076 1.00 006 098 

07[3= .00 124 703 071 2= .01912 741 

.-. sin0= .19 434 373 .-. cos0 = .98 093 357 

The sine is correct to five places and the cosine to four places. 

84. Simpson's method of computing a trigonometric table is the 
following : 

Suppose the table is to be at intervals of V. 
Putting (n - 1) V for A and 1' for B in (1) and (3) of § 40, 
we obtain 

sin n V = 2 cos 1' sin (n - 1) 1' - sin (n - 2) 1'. (1) 
cos n V = 2 cos V cos (n — 1)1' — cos (n — 2) V. (2) 
Putting 30° for A in (1) and (4) of § 40, we obtain 

sin (30° + B) = cos B - sin (30° - B). (3) 

cos (30° + B)== cos (30° - B) - sin B. (4) 

Calculate sin 1' and cos V as in § 82 or by the series in § 83. 

Then giving to n in (1) and (2) the values 2, 3, 4, etc., suc- 
cessively, we obtain the sines and cosines of angles up to 30° 
at intervals of V. 

E.g., when n = 2, (1) and (2) become 

sin 2 7 = 2 cos V sin V — sin 0' = 2 cos V sin 1', 
and cos 2' = 2 cos V cos V — cos V = 2 cos 2 1' — 1, .... 



118 PLANE TRIGONOMETRY 

To obtain the sines and cosines of angles from 30° to 45°, 
we give to B in (3) and (4) the values 1', 2', 3', etc., succes- 
sively, making use of the results previously obtained by (1) 
and (2). 

E.g., when B = 1', (3) and (4) become 

sin 30° V = cos V - sin 29° 59', 
and cos 30° V = cos 29° 59' - sin 1', • • .. 

85. The hyperbolic functions. One form of the exponential 
series is 

/y»* /y»o /V*^ /y»5 

. tAj *Aj %L/ tAj 

e ^ 1 + * + {2 + {3 + jI + [5 + -' < d > 

where e is the base of natural logarithms. 

For the proof of (d), see § 95 in Taylor's Calculus, or § 326 
in the College Algebra. 

Putting x = 1, we obtain e = 2.718281. 

The functions (e x — e~ x )/2 and (e x + e~ x ) /2 are found to have 
properties analogous to those of sin x and cos x, and to have 
to the hyperbola relations analogous to those which sin x and 
cos x have to the circle ; for these reasons the first has been 
named the hyperbolic sine of x (sink x), and the second the 
hyperbolic cosine of x (cosh x). Thus we have 

sinh x = (e x — e~ x ) /2, "] 

\ r37~i 

coshx = (e x + e- x )/2.{ 

Following the analogy of the trigonometric functions, we 
define the ratio of sinh x to cosh x as the hyperbolic tangent of 
x (tanh x). 

Thus, tanh x = ^^- = *] ~~ *~* - [38] 

cosh x e x + e~ x L J 

The reciprocals of tanh x, cosh x, sinh x are called respec- 
tively the hyperbolic cotangent of x (coth x), the hyperbolic 
secant ofx (sech x), and the hyperbolic cosecant ofx (csch x). 



HYPERBOLIC FUNCTIONS 119 

86. The relations between the hyperbolic functions are in gen- 
eral analogous to, and sometimes the same as, the corresponding 
relations between the trigonometric functions. 

E.g., (e x + e-*) 2 /4 - (e* - e-*) 2 /4 = 1. 
.-. cosh 2 x — sinh 2 x = 1. 

Again, cosh (— x) = = cosh x, 

g — X pX 

and sinh (— x) = = — sinh x. 

V ' 2 

e 2x _ e -2x e x _ g-x gx + g-x 

Also, sinh 2 x = = 2 

2 2 2 

.-. sinh 2 x = 2 sinh x cosh x. 

Again, sinh x cosh y = (e x — e~ x ) (ev + e--v)/4 

= ( e x + y _ g-ar + y + eac-y _ g—a:-y)/4 ? (1) 

and cosh x sinh ?/ = (e x + e _a: ) (e? — er-y) /4 

= (g* + */ _|_ e -x + y _ gx-y _ e~ x -y)/\. (2) 
Adding (1) and (2), we obtain 
sinh x cosh y + cosh x sinh y = [e* + ^ — e — ( x + ^)] / 2 

= sinh (x + y). 
Example. Prove sinh (x — y)= sinh x cosh y — cosh x sinh y. 
cosh (x + y) = cosh x cosh ?/ + sinh x sinh y. 
cosh (x — y) = cosh x cosh ?/ — sinh x sinh y. 
The notation for the inverse hyperbolic functions is the same as that 
for the inverse trigonometric functions. 

EXERCISE XXVin 

Prove each of the following identities : 

1. tanh 2 x + sech 2 x = 1. 2. cosh 2 x + cosh 2 x = 1. 

tanh x + tanh y 

3. tanh (x + ?/) = 

1 -f- tanh x tanh y 

4. sinh 3 x = 3 sinh x + 4 sinh 3 x. 

5. cosh 3 x = 4 cosh 3 x — 3 cosh x. 

6. sinh-^x = cosh-Wl + x 2 = tanh- 1 (x/ Vl + x 2 ) . 

x -\-it 

7. tanh -1 x 4 tanh— 1 2/ = tanh -1 

1 +xy- 



CHAPTER VIII 



COMPLEX NUMBERS. DE MOIVRE'S THEOREM 



B 



/ 

/ 
/ 

/ 

/ 
/ 

I - - 1 




S 

\ 
\ 
\ 
\ 
\ 

+ 1 * ■ 1 


i 
\ 





/ 


\ 

V 

\ 


1 

> 


/ 

/ 
/ 
J 

/ 



87. The quality units, ± l, ± V— 1. Let ABA'B 1 be a 

circle whose radius OA is 1 unit in length. 
Then if OA = the unit +1; OA 1 = the unit ~1. 
Now OA x (~1)= OA'; that is, if OA is multiplied by ~1, 

OA is reversed in direction and 
becomes OA'. Suppose that OA 
reverses its direction by revolv- 
ing about in the plane of the 
figure; then, as ~~1 =( + V— l) 2 
HJ or ("V^T) 2 , it follows that OA 
will be reversed in direction if it 
is multiplied twice in succession 
by either + V— 1 or ~"V— .1. 
Hence it may be assumed that 
the effect of V— 1 or ~~V— 1 
as a multiplier will be to revolve 

OA through it/ 2. Suppose that + V— 1, as a multiplier, 

revolves OA in the positive direction, or counter-clockwise ; 

then ~~V— 1 will revolve OA in the negative direction, or 

clockwise. 

That is, OA X + V^T = OB, OA X ~~ V^T = OB 1 . 

Putting + 1 for OA and assuming the commutative law, we 

have . , 

OB = V-l, OB 1 = ~ V- 1. 

For brevity we shall denote V— 1 by i and "~V— 1 by ~~i. 

The quality units + 1 and ~1 are often called real units, and 

arithmetic multiples of these units are called real numbers. 

120 



B' 

Fig. 55 



QUALITY UNITS 121 

The quality units i and ~i are called imaginary units, and 
arithmetic multiples of these units are called imaginary 
numbers. 

Observe that +i and — i or +1 and — 1 include both the idea of the 
arithmetic 1 and that of oppositeness to each other. 

Since +1 x+l = +lor— 1 x _ 1 = +1, +1 or — 1 is its own reciprocal. 

Since +ix~i = +l, +i and — t are reciprocals of each other. 

Hence i and — t are 6o^ opposites and reciprocals of each other. 

Since (— £) 2 = i 2 = — 1, £/ie square of either imaginary unit is — 1. 

Also, ( + i) 3 = — i an( I ( — ^) 3 = *> that is, £Ae cw&e o/ ei^er imaginary 
unit is equal to the other. 

Again, (~i) 4 = i 4 = + 1 ; that is, the fourth power of either imaginary 
unit is +1. 

88. A directed line or a directed force is a line or force whose 
value includes not only its magnitude but also its direction. 
E.g., OA, OB, OA', OB', A 7 !, BB', in fig. 55, are directed lines. 

A directed line is often called a vector. 

Two directed lines or forces are equal when they have the 
same length or size and the same direction. 

Hence, if two vectors are equal and their origins coincide, 
their ends also will coincide. 

If a directed line or force is parallel to one of two perpendicular lines, 
as AOA' and BOB' in fig. 55, we can express both its magnitude and its 
direction by a real or an imaginary number. To enlarge our number 
concept so that we can, by the sum of a real and an imaginary number, 
express the magnitude and direction of any directed line or force w T hich 
is parallel to any line whatever in a given plane, as A OB in fig. 55, we 
need the principle of vector addition given below. 

89. In fig. 55, we have 

OA' + A 7 ! = OA, OB +BB 1 = OB', 

OA + IX' = OA', OB' + WB = OB. 

Each of these identities illustrates the following law of 
vector addition : 



122 



PLANE TRIGONOMETRY 




Fig. 56 



If the end of one vector is the origin of a second vector, the sum 
of these two vectors in its simplest form is the vector extending 
from the origin of the first vector to the end of the second vector. 

Thus, in its simplest form, the sum of the vectors AB and 
BC in fig. 56 is the vector AC. 

That is, AB + BC = AC. (1) 

The meaning of (1) is that transference from A to B fol- 
lowed by transference from B to C is 
identical in result with transference from 
A to C. 

Or, a point P moving from A to C 
along AC goes the distance AB in the 
direction AB plus the distance BC in the 
direction BC. 
Again, if two directed forces are represented by the vectors 
AB and BC, by the parallelogram of forces we know that 
their resultant (i.e. their sum in its simplest form) is repre- 
sented by the vector AC. 

Identity (1) is absurd when, as in Geometry, we consider 
only the length of lines ; it becomes true when, and only when, 
the value of each line includes both its length and its direction. 

90, Complex numbers. If a and b are real numbers, then 
a + ib is called a complex number, i.e. a complex number in 
the common form is the sum 
of a real and an imaginary 
number. 

To construct a + ib, lay off 
OM = a and MP = ib ; then 
a + ib = OM + MP = OP. 

Thus the vector OP repre- 
sents the complex number 
a + ib, or a ■+■ ib is the numerical measure of the vector OP. 




COMPLEX NUMBERS 123 

Take OH equal to 1 unit in length and draw HR _L OM. 
Let r = the arithmetical value of Va 2 + b 2 . (1) 

Then, from the right triangle OMP, we know that 

r = the number of units in OP. 

From the similarity of the triangles ORH and OMP, we 
have 

OR = a/r, RH=ib/r. 

Hence OH = OR + RH = a/r + ib/r. 

That is, a/r + i7>/r denotes 0#, the unit vector of OP. 
r is called the arithmetic value, or modulus, of the complex 
number a + ib, and a/r + ib/r its quality unit. 

If Z ATOP = <£, then a/r = cos <£, 6/r = sin <£. (2) 

.*. a/r + ib/r = cos <f> + i sin <£. (3) 

.'. a + ib = (a/r + ib/r) r = (cos <j> + i sin </>) r. (4) 

(cos <£ + i sin <£) r is the trigonometric form of a complex 
number in which cos <£ + i sin <£ is the quality unit and r the 
modulus. 

Either (a/r + ib /r)r or (cos <£ + i sin <f>)r in (4) is called 
the type form of the complex number a + ib. 

Hence a complex number in the type form is an arithmetic 
multiple of a quality unit. 

The angle <£ is called the angle, or amplitude, of the complex 
number a + ib. Between — it and + ir there is one, and only 
one, value of <£ which will satisfy equations (2). This value 
of cj> is called its principal value. 

If <j>' denotes the principal value of <£, the general value of 
<£ is 2 nir + <£', where n is any integer. Thus the angle, or 
amplitude, of a complex number is many-valued. 

When b = 0, a + ib = a, a real number ; 
when a = 0, a + ib = z&, an imaginary number. 



124 PLANE TRIGONOMETRY 

Thus a + ib includes reals and imaginaries as particular 
cases. 

Ex. 1. Reduce the complex number — 1 — V — 3 to the trigonometric 
type form. 

Here a = — 1, b = — V3. 

.-.r=Vl +3 = 2. by (1) 

Hence cos = — 1/2, sin = — V3/2. by (2) 

Since cos and sin are both — , is in the third quadrant ; hence 
its principal value is — 2 it/ 3. 

... _i _VZ~3 = [cos(-2 7r/3) + £sin(-2 7r/3)] -2. 

Hence the arithmetic value, or modulus, of — 1 — V— 3 is 2, and its 
quality unit is represented by a unit vector which makes the angle 
— 2 it/ 3 with the vector representing +1. 

Ex. 2. Reduce the complex number 2 — V— 5 to the algebraic type 
form. 

Here a = + 2, b = - V5 ; .-. r = 3. 

.-. 2 - V^5 = p/3 + i(- V5/3)] 3. 

Ex. 3. Reduce the unit sin — i cos to the trigonometric type form. 

sin - i cos = cos (0 - 90°) + i sin (0 - 90°). 
Thus the angle of the quality unit sin — i cos is — 90°. 

EXERCISE XXIX 

Represent graphically and reduce to the type form : 

1. 1+V31. 5. -^/3 + i 9. _6-i8. 

2. 1-V^l. 6. -V3-i 10. -V6-*VH. 

3. — 1 — i. 7. 3 — ii. 11. sin0 + icos0. 

4. - 1 + V3~3. 8. - 3 + i 2. 12. - sin + i cos 0. 
13. By constructing the sum in each member, show that 

(a + ib) + (x + iy) = (a + iy) + (se + i&) = (a + x) + i (6 + y), 

and thus prove geometrically that the commutative and associative laws 
of addition hold true for complex numbers. 



GENERAL QUALITY UNIT 



125 



91. General quality unit cos <)> + i sin <}>. 

Let ABA' be a circle whose radius is 1, and let OA = the 
unit +1. 




Then if <f> = A OP, cos <f> + i sin <f> = OP ; 

i£ cf> = A OP/, cos <£ + i sin <£ = OP/ ; 

if cf> = A OP x , cos <£ + i sin <£ = OP 1 ; 

if <£ = A OP 1 , cos <£ + i sin <£ ee OP'. 

That is, the quality unit cos <f> + i sin <£ is represented by 
the unit vector which makes the angle <f> with the vector repre- 
senting + 1. 

The following important particular cases of cos <£ + i sin <£ 
should be carefully noted : 

When <j> = 0, cos <j> + i sin <j> = cos + i sin = + 1. 

When </> = ±tt/2, 

cos cj> + i sin<£ = cos(± 7r/2) + i sin(± 7r/2)= ± i. 
When <£ = ± 7r, 

cos <£ + i sin <£ = cos (± 7r) + £ sin (± ir) = ~~1. 

These results are evident also from the figure. 



126 PLANE TRIGONOMETRY 

92. Product of quality units. By the distributive law of 
multiplication, we obtain 

(cos fa + i sin fa) (cos fa + i sin fa) 
= cos fa cos fa + i cos fa sin <f> 2 + i sin fa cos fa + i 2 sin fa sin <£ 2 
= cos fa cos fa — sin fa sin <£ 2 + i (sin ^ cos <£ 2 + cos fa sin <£ 2 ) 
= cos (fa + <£ 2 ) + i sin (^ + fa). § 34 

Hence, in general, we have 

(cos <!>! + i sin § x ) (cos cj> 2 -h i sin (j> 2 ) • • • (cos c|> n + i sin <|> n ) 
= cosfti + <|> a + • • ■ + cj> n ) + i sin^ + <|> 2 + • • • + c|> n ). [39] 

That is, the product of two or more quality units is a quality 
unit whose angle is the sum of the angles of the factors. 

Observe that, to multiply cos <pi + i sin X by cos 2 + i sin 2 , we add 
the angle 2 of the multiplier to the angle <pi of the multiplicand. 

Again, to obtain the multiplier cos 2 + i sin 2 from the primary unit 
cos + i sin 0, we add the angle 2 of the multiplier to the angle of the 
primary unit. 

Hence the product is obtained by doing to the multiplicand just what 
is done to the primary unit to obtain the multiplier. 

93. De Moivre's theorem. 

(i) If <f> = fa = fa = ... = fa, 

then, from [39] of § 92, we obtain 

(cos <£ + i sin fa) n = cos n<\> + i sin nfa (1) 

That is, the nth power of a quality unit is a quality unit 
whose angle is n times the angle of the base. 

(ii) Taking the nt\i root of each member of (1), we obtain 

(cos ncf> + i sin ?i<j>) 1/n = cos <f> + i sin fa 
Putting <j> for ncf> and therefore <j> /n for <f>, we have 

(cos <£ + i sin fa) l/n = cos (<l>/n) + i sin (<j>/n). (2) 



DE MOIVRE'S THEOREM 127 

Let s be a positive integer, then, from (2), we obtain 
(cos <f> + i sin cf>) s/n = [cos (<f>/n) + i sin (<t>/n)y 

= cos (s<f>/n) + i sin (s<f>/n). (3) 
(iii) By §92, 

(cos <£ + isin<£)[cos(— 4>) + i sin(— <£)] = cos + i'sinO = + l. 
Hence 

cos (— cj>) + i sin (— <£) is the reciprocal of cos <f> + i sin <£. 
That is, (cos <£ + i sin <£) _1 = cos (— <£) + i sin (— </>). (4) 
Let^> be any positive integer or fraction. Then, from (4), 
(cos <f> + i sin </>) -p = [cos (— <£) + i sin (— <£)] p 

= cos (— pcfr) + £ sin (— ^<£). (5) 

From (1), (3), and (5) it follows that for any commensura- 
ble real value of n, we have 

(cos <|> + i sin c(>) n = cos ncj> + i sin n<(>. [40] 

Formula [40] is called De Moivre's theorem. 

Cor. 1. From (2), cos (<£/n) + i sin (<£/n) is one of the nth 
roots of cos <f> + i sin <j>. 
Cor. 2. By § 28, 

cos (— <j>) + i sin (—<£) = cos <£ — i sin <£. 

Hence, by (4), cos <j>— i sin cf> is the reciprocal of cos <f> + i sin <£. 

That is, the conjugate quality ?(?iits cos <£ + i sin <£ <mc? 
cos </> — i sin </> are reciprocals of each other. 

In fig. 58 observe that OP and OP 1 represent reciprocal, or 
conjugate, quality units, while OP and OP' represent opposite 
quality units. When OP coincides with OB, OP 1 and OP' both 
coincide with OB'-, this illustrates that the reciprocal, or con- 
jugate, of i is also the opposite of i. When OP coincides with 
OA or OA', OP x does also; that is, either + 1 or ~1 is its own 
reciprocal. 



128 PLANE TRIGONOMETRY 

94. To divide by the quality unit cos <f> + i sin <£, we multiply 
by its reciprocal cos (— <£) + i sin (— <£), or cos <£ — i sin <£. 

^ „ cos 01 + ^ sin 0i , . . v r , x . . . 

Ex. 1. — = (cos 0i + i sin 0i) [cos ( — 2 ) + i sin ( — 2 )] 

cos 02 + i sin 2 

= cos (0i — 2 ) + i sin (0i — 2 ). 

_ _ (cos 4- i sin 0) 5 , . . ,,. . . „._ 

Ex. 2. ^- — — ^- = (cos + z sm 0) 5 (cos + i sin (9) 7 

(cos0-isin0)7 v * T ^ ; v T ' 

= (cos 5 + i sin 5 0) (cos 7 + £sin 7 0) 

= cos (5 + 7 0) + i sin (5 + 7 0). 



EXERCISE XXX 

Prove each of the following identities : 

1. (cos + z sin 0) 3 = +1, [cos(2?r/3) + £sin(2 tt/3)] 3 = +1, 
[cos(4 7zr/3)+£sin(4 7zr/3)] 3 = +1. 

Hence each of these three quality units is a cube root of + 1. 

2. [cos(7r/5)±ism(7r/5)] 5 =-l, [cos(3tt/5)± i sin(3^r/5)] 5 =-1, 
(cos it ± i sin it) 5 = ""1. 

Observe that cos it 4- i sin it = cos it — i sin it. 

3. [cos (it/ 6) ± i sin (it/6)f = [cos(tt/2)± isin(^r/2)] 6 

= [cos (6 it/6)±i sin (5 ?t/6)f = -1. 

4. What is proved by examples 2 and 3 ? Illustrate the meaning of 
examples 1, 2, 3 by vectors. 

Express as a quality unit each of the following : 

e (cos + i sin 0) 6 (cos — i sin 0) 3 /2 

5. b. • 

(cos — i sin 0) 5 (cos0 — zsin0) 5/3 

(cos + i sin 0) (cos + i sin 0) 
(cos /3 -f i sin /3) (cos y + £ sin 7) 

R [cos (7T/6) - isin(7T/6)] n / 2 (cos + i sin 0) 4 

[cos (it/6) + i sin (tt/6)] 1 /2 ' ' (sin + i cos 0) 5 ' 



COMPLEX NUMBERS 129 

10. By De Moivre's theorem prove identities (1) and (2). 

n(n — 1) 
cos n0 = cos" cos n_ 2 <p sin 2 

n(n- 1) (n - 2) (n - 3) 
+ — * ^- — cos *- 4 sm 4 . (1) 

n(n - l)(n -2) _ . _ 

sin ?i0 = ncos n_1 sin '— cos n ~ 3 <p sin 3 

L§ 

>i (w - 1) (n - 2) (n - 3) (n - 4) _ . _ 
+ — — — — — ^ -cos»- 5 0sm 5 . (2) 

cos n0 -f i sin n0 = (cos -f i sin 0) n 

= cos n + -i - cos"- 1 sin + i 2 n ^ n ~ * cos n ~ 2 sin 2 
1 \2 

+ t» n(n " 1)(w " 2) cos"--g0Bin»0 + • • .. (3) 

Substituting for i 2 , i 3 , • • • their values, and equating the real parts 
and the imaginary parts in (3), we obtain (1) and (2). 

95. Products and quotients of complex numbers. Multiplying 
by a quality unit affects only the quality of the product, and 
multiplying by an arithmetic number, or modulus, affects only 
the size of the product ; whence the result is evidently inde- 
pendent of the order of these operations. Therefore the com- 
mutative and associative laws of multiplication hold for the 
moduli and quality units of complex numbers. 

Hence we have the following theorems : 

(i) The product of two or more complex numbers is equal to 
the product of their quality units into the product of their moduli. 

(ii) The quotient of two complex numbers is equal to the quo- 
tient of their quality units into the quotient of their moduli. 

(iii) The mth power of any complex number is equal to the 
mth power of its quality unit into the mth power of its modulus, 
where m is any real number. 



130 PLANE TRIGONOMETRY 

96. The quality unit cos cj> + i sin </> has q, and only q, unequal 
qth roots. 

If n is any integer, then, by § 21, we have 

cos <j> + i sin <£ = cos (2 ?i7r + <£) + i sin (2 W7r + <£). 
.-. (cos<£ + isin<£) 1/<z = [cos(2^tt + </>)+ i sin (2 wtt + <£)]* « 

= cos — - + * sm (1) 

q q 

If in (1) we give to n the values 0, 1, 2, 3, 4, • • •, q — 1, in suc- 
cession, we obtain the following q qth roots of cos <$> + i sin <£ : 

n = 0, cos (<l>/q) + i sin (<f>/q), 

2tt + d> . . 2?r + d> 
n = 1, cos + fc sin ? 

4?r + <£ . 4tt + <£ 
ra, = 2, cos + % sm ? 



(2) 



The angles of any two of the roots in (2) differ by less than 
2 7r; hence no two of these angles can have equal cosines and 
equal sines. Therefore the q qth. roots in (2) are unequal. 

If we give to n the values q, q + 1, q + 2, • • •, or — 1, — 2, 
— 3, • • •, we obtain ^th roots equal to those in (2). 

Hence cos <£ + i sin cj> has q, and only q, unequal qth roots. 

97. Any complex number has q, and only q, unequal qth roots. 
[(cos <£ + i sin <j>)r~\ 1/q = (cos <£ + i sin <^) 1/ V /5 . (1) 

The quality unit cos <£ + i sin <£ has q, and only q, unequal 
^th roots, and the arithmetic number r has only one qth root ; 
whence the second member of (1) has q, and only q, unequal 
values. 

Hence to find the q qth. roots of any complex number, reduce 
the number to the type form, find the q qth roots of its quality 
unit, and multiply each by the ^th root of the modulus. 



ROOTS OF COMPLEX NUMBERS 131 

Observe that an algebraic number has q unequal qth roots because its 
quality unit has q unequal gth roots. 

Ex. 1. Find the three cube roots of — 27. 
- 27 = ( - 1) • 27 = (cos it + i sin it) ■ 27. 
.-. V- 27 = (cos7T + ismx) 1 '* -3. 

, ... v1/q 2tt7T + It . . 2ll7t + It 

(cos 7t -\- ism7t) 1/B = cos 1- 1 sin ; 

o o 

when?i = 0, = cos(*r/3) + i sin (tt/3) = 1/2 + i V3/2; 

when n = 1, = cos tt + i sin 7r = — 1 ; 

when n = 2, eecos(5tt/3) + isin(5^r/3) = 1/2 -i V3/2. 



Hence V- 27 = -3, (1/2 ± i V3/2) • 3. 

Ex. 2. Find the four fourth roots of 8 + 8 VZ3. 

8 + 8 V^ =(1/2 + i V3/2) • 16 = [cos (tt/3) + i sin (tt/3)] - 16. 
... V8 + 8 V^3 = [cos(tt/3) + i sin (tt/3)] 1 ' i - 2. 

r / /o\ i • • i /o\m/4 2717T + ^/3 . . 2 717T + TT/3 

[cos(7T/3) -f zsm(7r/3)] 1/4 = cos h zsm — ; 

4 4 

when n = 0, = cos (tt/12) -f isin (?r/12); 

when n = l, = cos(7 7r/12) + isin (7 7T/12); 

when n = 2, = cos (13 it / 12) + i sin (13 it / 12) 

whenn = 3, = cos (19 tt/12) + isin (19 it / 12). J 



(1) 



Multiplying each of the four units in (1) by 2, we obtain the four 
required fourth roots. 



EXERCISE XXXI 

Find alljthe values of : 

1. li/s. 5. 16 1 / 4 . 9. (4V3 + i4)!/ 3 . 

2. (-I) 1 / 3 . 6. 32^/5. 10. (l _ i V3)V4. 

3. I 1 / 6 . 7. (-243) 1 / 5 . 11. (l+V31)i/6. 

4. (-1) 1 / 6 . 8. (-0 1/6 - 12. (V3-VTl)2/5 

13. Solve the equation x 4 - x 3 + x 2 - x + 1 = 0. (1) 

Multiply by (x + 1), x 5 + 1 = 0, or x = (- I) 1 / 5 . 



132 PLANE TRIGONOMETRY 

Multiplying by x + 1 introduces the root — 1, the other four of the 
five fifth roots of — 1 are the roots of equation (1). 

14. Solve the equation x 12 — 1 = 0. 
Factor, (x 3 - 1) (x 3 + 1) (x 3 - i) (x 3 + i) = 0. 

The twelve roots, l 1 /*, (- l) l / s , z 1 / 3 , (- i) 1 / 3 , are readily found by 
De Moivre's theorem. 

15. Solve the equation x 7 + x 4 + x 3 + 1 = 0. 
Factor, (x 4 + 1) (x 3 + 1) = 0. 

16. Solve the equation x 7 + 1 = 0. 

98. Exponential form for cos <j> + i sin c|>. If in series (d) of 
§ 85 we replace x by i<j>, we obtain the series (1). 

1 + u 4 W + (i+Y 4. (^) 4 + W 4 n ^ 

S ( 1 -f + f-f + -) + '(^"l + l"f + '-) §87 
= cos <£ + i sin <£. by (b), (c), § 83 

That is, series (1) is equal to a quality unit whose angle is <j>. 
If, as is suggested by (d), we define e i(t> as an exponential 
symbol for the series (1), we have 

c!* = cos <£ + i sin <£. (2) 

That is, €** is an exponential symbol for a quality unit whose 
angle is <|>. 

Substituting — <f> for <£ in (2), we obtain 

c"** = cos <£ — i sin <f>. (3) 

From (2) and (3), e i0 and c~** denote reciprocal quality units. 
From (2) and §§92 and 94, it follows that 

€ «f>l . £ #2 . £ i<f>3 = e K^>l + 02 + <l>3) € *0 . € -«0 = e «(0-0) 

C 10 _i_ € t0 = € 10 . C -10 = £ ?(0-0) ( € l( />) 2 = C i2 0. 



THE EXPRESSION c'* 133 

That is, € i(p obeys the fundamental laws of exponents. 
The unit of cj> is the radian. Why ? 
Putting 1 for <f> in (2), we obtain 

e 1 ' = cos 1 + i sin 1. 

That is, e* denotes a quality unit whose angle is a radian. 

Example. What does e _i ' denote ? e° ? 

Observe that for all values of </> the arithmetic value of e l * is 1. 

Hence, as <p varies from — <x> to -f <», e"*> varies in quality only. 

Therefore, e in e^ cannot have the meaning it has in e x . 

Observe that the numerical measure of any directed line or force in a 
given plane can be expressed in the form e 1 '* • a, where a is an arithmetic 
number which gives its length or size and is some angle between — it 
and it which gives its direction. 

Adding and subtracting (3) and (2), we obtain 

cos <£ = > sm <j> = — — (4) 

The values in (4) are known as Euler's exponential values 
of cos 4> and sin <£. 

Compare (4) with [37] in § 85. 



CHAPTEE IX 

MISCELLANEOUS EXAMPLES 

EXERCISE XXXII 

Express all the angles which are coterminal with : 

1. 45°. 2. 132°. 3. -35°. 4. -100°. 5. 7t/6. 

Find all the other trigonometric ratios of A when : 

6. sin 4 = 4/7. 8. cos4 = -3/8. 10. sec4 = 7/4. 

7. tan4 = 3/2. 9. cot4 = -7/5. 11. csc4 = -5/4. 

12. In what quadrant is A in each of the examples 6-11 inclusive ? 
Construct A in each. 

In terms of a function of an angle less than 45° express : 

13. sin 94°. 16. cot 320°. 19. cos (- 175°). 

14. cos 128°. 17. sec 190°. 20. tan (-200°). 

15. tan 215°. 18. sin (-75°). 21. cot (-300°). 
In terms of each of the other functions of A find the value of : 

22. sin A. 24. tan A. 26. sec 4. 

23. cos A. 25. cot A. 27. csc4. 

Identities 

Prove each of the following identities : 

28. (tan A + cot A) sin A cos A = 1. 

29. (sec A — tan A) (sec A + tan A) = 1. 

30. (esc A — cot A) (esc A + cot A) = 1. 

31. (sin B - cos J5) 2 = 1 - 2 sin B cos B. 

32. sin 5 + cos 5= V2cos(£- rt/4). 

33. sin£-cos.B = -V2cos(jB+ tt/4). 

134 



IDENTITIES 135 

34. sin (A -f tt/S) + sin (A - tt/3) = sin A. 

35. cos (A + 7T/6) + cos (A - 7t/6) = V3 cos A. 

36. (cot J. + tan B) / (tan A + cot E) = cot A tan 5. 

37. 1 - tan 4 A = 2 sec 2 J. - sec 4 ^4. 

38. sec J?/ (1 + cos B) = (tan B - sin £) /sin 3 B. 

39. sec 2 ^4 esc 2 A = tan 2 ^4 + cot 2 A + 2. 

40. tan 5 + sec 5 = tan (J5/2 + tt/4). 

41. (1 + tan £) / (1 - tan 5) = (cot B + 1) / (cot 5 - 1). 

42. sin 4 / (1 + cos A) + (1 -f cos ^4) /sin ^4 = 2 esc ^. 

43. sec -3 A — sin 3 ^4 = (cos A — sin A) (1 + sin ^4 cos A). 

44. (sin ^4 cos B -f cos ^4 sin i?) 2 + (cos J. cos B — sin ^4 sin B) 2 = 1. 

45. cot A — tan J. = 2 cot 2 ^4. 

46. sec 2 ^4 = sec 2 ^4/ (2 -sec 2 ^4). 

47. 2sec 2 A = sec(A + 7t/A)sec(A - 7T/4). 

48. sin 2 A = 2 tan A /{I + tan 2 ^4). 

49. 2 sin J. + sin 2 .4 = 2 sin 3 A / (1 - cos ^4). 

50. Find sin (A + 5 + C) in terms of the sine and cosine of A, B, C. 
Applying formula [7] twice and [8] once, we obtain 

sin (A + B + C) = sin (A + B) cos C + cos (^4 + B) sin C 
= (sin A cos 5 + cos A sin J5) cos C 

+ (cos ^4 cos B — sin A sin I?) sin C 
= sin ^4 cos B cos C + cos A sin I? cos C 

+ cos A cos .B sin C — sin ^4 sin B sin C. (1) 

If A, B, C are the angles of a triangle, sin (A + B + C) = sin 180° = 0. 
Hence, from (1), we obtain 
sin A cos B cos C + cos ^4 sin J5 cos C -f cos J. cos I? sin C = sin A sin 5 sin C 

51. Applying formula [8] twice and [7] once, prove 

cos (A -f B + C) = cos A cos B cos O — cos A sin J5 sin C 

— sin A cos 2? sin C — sin ^4 sin B cos C. (2) 



136 PLANE TRIGONOMETRY 

52. Find tan (A + B'+ C) in terms of tan A, tan JB, tan C. 
Applying formula [11] three times, we obtain 

tan (^4 + B) + tanO 



tan (A + B + C) = 



1 - tan (A + B) tan (7 
tan J. + tan B 



1 — tan A tan 5 



+ tan C 



„ tan ^1 + tan 5 _ 

1 tan C 

1 — tan A tan 5 



tan A + tan I? + tan C — tan J. tan B tan (7 

__ . ^j j 

1 — tan ^i tan 5 — tan A tan (7 — tan B tan C 

If A, B, C are the angles of a triangle, tan (A + B + C) = tan 180° = 0. 
Hence, from (3), 

tan A + tan 5 + tan C = tan ^4 tan B tan 0. 

53. Putting A = B = Cin (1), (2), (3) of examples 50, 51, 52, prove 

sin 3 A = 3 sin ^L (1 - sin 2 A) - sin 2 ^i 

= 3 sin A — 4 sin 3 A, (4) 

cos 3 A = 4 cos 3 A — 3 cos J., (5) 

_ . 3 tan A — tan 3 A 

tan 3^1 = (6) 

l-3tan 2 ^i v ' 

Identity (5) is useful in solving cubic equations. See example 112. 

54. Writing SA = 2A + A, prove (4), (5), (6) in example 53 by using 
[7], [8], [11], [13], [14], and [15]. 

55. Substituting for 3 A in (4) and (5) of example 53, we obtain 

sin = 3 sin (0/3) - 4 sin 3 (0/3), 
cos = 4 cos 3 (0/3) - 3 cos (6/3). 

Prove each of the following identities : 

56. sin 4 A = 4 sin A cos A — 8 sin 3 J. cos -4 

= 8 cos 3 A sin A — A cos J. sin J.. 

57. cos4^4ee 1 - 8 cos 2 ^4 + 8 cos 4 J. 

= 1 - 8 sin 2 A + 8 sin 4 A. 

58. cos 780° =1/2. 60. cos 2550° = y/3/2. 

59. sin 1485° = V2/2. 61, sin (- 3000°) = - V3/2, 

62. tan ( - 2190°) = - y/3/S, 



EQUATIONS 137 

Equations 

In what quadrant is A in each of the following equations ? 

63. sin A cos A - - 2 /3. 65. sec A tan A = - 3. 

64. sin J. tan A = 4. 66. cot ^L + 3 sin A = 0. 

67. Solve the equation sin 2 = 2 cos 0. (1) 

Substituting for sin 2 its identical expression 2 sin cos 0, from (1) by 
Algebra we obtain the equivalent equation 

2 sin cos = 2 cos 0, or cos (sin — 1) = 0. 
From cos = 0, = nit ± 7t/2. § 70 

From sin 0=1, B = war +(- l)»*r/2. §69 

Hence ?i7r ± 7T/2 includes all the values of in (1). 

Solve each of the following equations : 

68. cos 2 = 2 sin 0. 82. 2 sin 2 x - 2 = - V2 cos x. 

69. cos = sin 2 0. 83. cos 2 ?/ + 2 sin 2 ?/ - f sin?/ = 0. 

70. sin = cos 2 0. 84. sin + sin 2 = 1 - cos 2 0. 

71. tan A tan 2 A = 2. 85. cos y — cos 2 ?/ = 1. 

72. cos J. + cos 2 A = 0. 86. sin (45° + 2) + cos (45° - z) = 1. 

73. cot J. tan 2 .A = 3. 87. sec 2 2 + 1 = 2 cosz. 

74. 4 cos 2 A + 3cosv4 = 1. 88. cos 2 2 = a (1 - cos 2). 

75. sin sec 2 = 1 . 89. tan 2 y tan y = 1. 

76. cot tan 2 = sec 2 0. 90. sec = 2 tan + 1/4. 

77. sin 2 = 3 sin 2 - cos 2 0. 91. sin-ix + sin- 1 (x/2) = 120°. 

78. sin + cos 2 = 4 sin 2 0. 92. sin- 1 z + 2 cos~ x z = 210°. 

79. sin20 = cos4 0. 93 tan- 1 ?/ + 2 cot" 1 ?/ = 135°. 

80. sec x + tan x = ± VS. 9 2 

94. tan- 1 -— - = 60°. 

81. tan x + 2 V3 cos x = 0. 1 - 2 2 

95. tan- 1 z + tan- 1 2 2 = tan- 1 3 V3. 

96. tan x + tan 2 x = 0. 



138 PLANE TRIGONOMETRY 

97. tan 2 x + cot 2 x = 10/3. 99. sin A + cos A = sec 4. 

98. 4 cos 2 4- 6 sin = 5. 100. sin(0 + 30°) sin (0 - 30°) = 1/2. 

Systems of Equations 

101. Solve for r and the system 

r sin = a, (1) 



r cos = b. (2) r 
Divide (1) by (2), tan = a/b, or = tan- 1 (a/6). 

Square (1) and (2) and add, 

r 2 (sin 2 + cos 2 0) = a 2 4- b 2 . 
.-. r = Va 2 + 6 2 . 

102. Solve for r, 0, and the system 

r cos sin = a, (1) 1 

r cos cos = 6, (2) L (a) 

r sin = c. (3) J 

Divide (1) by (2), tan = a/b, or = tan- 1 (a/b), (4) 

Square (1) and (2) and add, 

r 2 cos 2 = a 2 4 6 2 . (5) 

From (5), r cos = Va 2 + ft 2 . (6) 

Divide (3) by (6), tan = c/Va 2 + 6 2 , or = tan- 1 (c/ Va 2 + 6 2 ). 
Square (3) and add (5), 

r = Va 2 + 6 2 + c 2 . (7) 

103. Solve for x and y the system 
sin x + sin y = a, (1) 



:} 



(a) 

cos x + cos y = b. (2) 

By § 40 we obtain from system (a) the equivalent system (b). 

(b) 



2 sin i (x 4 y) cos J (x — y) = a, (3) 1 



2 cos f (x + y) cos J (x — y) = b. (4) j 

Divide (3) by (4), tan ±(x + y) = a/b. (5) 

Hence sin \ (x + y) = ± a / Va 2 4- 6 2 . (6) 

Substituting the value of sin \ (x 4- y) in (3), we obtain 

cos i(x-y) = ± Va 2 + 6 2 /2. (7) 



SYSTEMS OF EQUATIONS 139 

From (5), z + y = 2 tan- 1 (a/b). (8) 1 

From (7), x-y = 2 cos- 1 (± Va 2 + 6 2 /2). (9) J ( °' 
Hence x = tan- 1 (a/6) + cos- 1 ( ± Va 2 + 6 2 /2), 

and 2/ = tan- J (a/b) - cos~ x ( ± Va 2 + & 2 /2) . 



104. Solve the system 



sm x — sin 
cos x — cos 






105. Solve for x and ?/ the system 

x sin ^4 + y sin 5 = a, (1) ^ 

r (a) 

x cos ^L + ?/ cos B = 6. (2) j 

Since (1) and (2) are each linear algebraic equations in x and y, 
system (a) is solved as a linear algebraic system. 

106. Solve for r and 6 the system 

r sin (d + A) = a,^ 

> (a) 

r cos (0 + 5) = b. J w 

By [7] and [8], from (a) we obtain the equivalent system (b). 

r sin 6 cos A + r cos sin J. = a, i 

> (b) 
r cos cos B — r sin 6 sin I? = 6. J 

Solve (b) as an algebraic linear system in r sin 6 and r cos as the 
unknowns. 

The resulting system can then be solved for r and as in example 101. 



107. Solve the system 

Ux-v\ = 2. fm 

(a) 



cos (x + y) + cos (x - y) = 2, (1) ^ 



sin(x/2) + sin(?//2) = 0, (2) J 
for values of x and y less than 2 it. 

By [8] and [10], from (1) we obtain the equivalent equation 
cos x cos y — sin x sin y + cos x cos y -f sin x sin y = 2, 
or cos x cos 2/ = 1. (3) 

From (3), cos x and cos y are both + 1 or both — 1. Why ? 

Hence x and y are both coterminal with or both coterminal with it. 



;} 



140 PLANE TRIGONOMETRY 

The solution x = 0, y = of (3) satisfies (2) ; also the solution x = n, 
y = — 7t, or x = - rf, y = 7t of (3) satisfies (2). 

Observe that either x = it, y = it, or x = - nr, y — - it is a solution of 
(3), but neither is a solution of (2). 

108. Solve for R and F the system 

W- Fsmh-Rcosh = 0,^ 
W- Fcos h - R sin h = 0. j 
Observe that this system is algebraic and linear in R and F. 

109. Eliminate from the system 

x = r(d — sin 0), (1)* 

y = r(l - cos0). (2), 
From (2), y — r vers 0, or = vers -1 (y/r). (3) 

From (2), cos = (r — ?/)/r. .-. sin = ± ^2ry — y 2 /r. (4) 

From (1), (3), (4), x = r vers- 1 (y/r) T V2 ry - y 2 . (5) 

110. Eliminate from the system 

a cos + b sin = c, (1) 

dcos0 + esin =/. (2). 

Solving the system for cos and sin 0, we obtain 

. « af — cd ce — bf 

sin = - , cos = -U . (3) 

ae — bd ae — bd 

Squaring the members of equations (3) and adding, we obtain 

(af - cd) 2 + (ce - bf) 2 . . , „ * 

-^ ^— — ^ J -?- = sm 2 + cos 2 = 1. 

(ae - bd) 2 

.-. (ae - bd) 2 = (af- cd) 2 + (ce - bf) 2 . 

111. Eliminate from the system 

a cos + b sin = c, 6 cos — a sin = d. 

112. Solve the cubic equation x 3 — 3px + <7 = 0. (1) 
Putting x = z/n, we obtain 

z 3 - 3pft 2 z + qn* = 0. - (2) 

Now by (5) in example 53, we have the identity 
cos 3 A = 4 cos 3 A — 3 cos A, 
or cos 3 .4 - (3/4) cos A - cos (3 A) /4 = 0. (3) 



;} 



CUBIC EQUATIONS 141 

Comparing identity (3) with equation (2), we see that cos A is a root of 
(2) when n and A satisfy the conditions 

3p?i 2 = 3/4, andgn 3 = - cos(3^L)/4. 
Hence n = l/(2Vp), 

and cos 3 A = — 4qn z = — q/(2p 3/2 ). (4) 

Observe that (4) can always be solved when p is positive and g/(2p 3/2 ) 
is arithmetically equal to or less than 1. 

If Ai is the principal value of A which satisfies (4), then the values 
A\ + 2?r/3 and A\ + 4 7T/3 also satisfy it. 

Hence the roots of equation (1) are 

cos Ai/n, cos (Ai 4- 2 tt/3) /n, and cos (A\ + 4 tt/3) /n, 
i.e. 2 Vp cos A±, 2 y/p cos(^i x 4- 2 7T/3), and 2VP cos (-4i 4- 4 7r/3). 

By Algebra, we know that the general cubic equation 
?/ 3 4- 3 ay 2 + by 4- c = 
can be transformed into one of the type (1) by putting y = x — a. 

113. Solve x 3 + 6 x 2 + 9 x + 3 = 0. 
Putting x = y — 2, we obtain y z — 3 y 4- 1 = 0. 
Putting y = z/n, we obtain z 3 — 3 ?i 2 z + n z = 0. 
Now cos 3 ^4. - (3/4) cos^. -(1/4) cos 3 A =0. 

Hence z = cos A, when n 2 = 1 /4 and n 3 = — cos 3-4/4, 

e.e. when ?i = 1/2, and cos 3 A = — 1/2 = cos 120°. 

.-. 3^1 = 120°, or ^Li = 40 o . 
Hence z = cos 40°, cos (40° + 120°), or cos (40° + 240°). 

.-. y — 2 cos 40°, 2 cos 160°, or 2 cos 280°. 
.-. x = - 2 + 2 cos 40°, -2 -2 cos 20°, or -2 4-2 cos 80°. 

Having given sin 15° = ( V3 - l)/(2 V2), and cos 15° = ( V3 4- 1)/(2 V2), 
solve each of the following equations : 

114. x 3 -24x-32 = 0. 116. 2x 3 -3x-l = 0. 

115. x 3 -6x 2 4-6x4-8 = 0. 117. x 3 4-3x 2 -l = 0. 

118. x 3 + 4 x 2 4- 2 x - 1 = 0. 



142 PLANE TRIGONOMETRY 

EXERCISE XXXIH 
Triangles 

1. Two towers are 3 mi. apart on a plain. The angle of depression 
of one, from a balloon directly above the other, is observed to be 8° 15'. 
How high is the balloon ? 

2. The shadow of a tree 101.3 ft. high is found to be 131.5 ft. long. 
Find the elevation of the sun. 

3. A rock on the bank of a river is 130 ft. above the water level. From 
a point just opposite the rock on the other bank of the river the angle of 
elevation of the rock is 14° 30 / 21". Find the width of the river. 

4. A rope 38 ft. long, when fastened to the top of a tree 29 ft. high, 
just reaches a point in the plane of the foot of the tree. Find the angle 
which the rope makes with the ground. 

5. A window in a house is 24 ft. from the ground. Find the inclina- 
tion of a ladder placed 8 ft. from the side of the building and reaching 
the window. 

6. A ladder 40 ft. long reaches a window 33 ft. high, on one side of a 
street. Its foot being at the same point, it will reach a window 21 ft. 
high on the opposite side of the street. Find the width of the street. 

7. A lighthouse 54 ft. high is situated on a rock. The angle of eleva- 
tion of the top of the lighthouse, as observed from a ship, is 4° 52', and 
the angle of elevation of the top of the rock is 4° 2'. Find the height of 
the rock and its distance from the ship. 

8. A man standing south of a tower, on the same horizontal plane, 
observes its angle of elevation to be 54° 16' ; he goes east 100 yd. , and 
then finds its angle of elevation to be 50° 8'. Find the height of the tower. 

9. A pole is fixed on the top of a mound, and the angles of elevation 
of the top and the bottom of the pole are 60° and 30° respectively. 
Prove that the length of the pole is twice the height of the mound. 

10. Given that the radius of the earth is 3963 mi., and that it sub- 
tends an angle of 57 / 2" at the moon. Find the distance of the moon 
from the earth. 



TRIANGLES 143 

11. Given that the radius of the earth is 3963 mi., and that it sub- 
tends an angle of 9" at the sun. Find the distance of the sun from the 
earth. 

12. Solve example 1 in § 57 by the principles of right triangles. 
The given parts are a side and two angles. 

In fig. 38 draw AH A. BC. 

In the right triangle AHB, compute the sides c and BH. 

Then compute c in the right triangle HCA. 

13. Solve the first four examples in Exercise XIX by the principles of 
right triangles. 

14. Solve example 1 in § 58 by the principles of right triangles. 
The given parts are two sides and an angle opposite one of them. 
In fig. 39 draw CH ± AB. 

Compute the sides CH and AH in A ABH 
Then compute B and HB in A HBC. 

15. Solve examples 1, 3, 5, and 7 in Exercise XX by the principles of 
right triangles. 

16. Solve the example in § 59 by the principles of right triangles. 
The given parts are two sides and their included angle. 

In A ABC draw BH ± CA. 
Compute CH and BH in A CHB. 
Compute A and c in triangle BHA. 

17. Solve the first four examples in Exercise XXI by the principles 
of right triangles. 

18. Solve example 6 in § 60 by the principles of right triangles. 
The given parts are the three sides. 

In the A ABC draw AH JL BC and let x = HC. 

: AH 2 = c 2 -(a- x) 2 . 
: c 2 - a 2 - x 2 + 2 ax. 
:(a 2 + b 2 -c 2 )/(2a). 

Whence HC and b are known in A HAC, and c and BH in A BAH. 

19. Solve the first four examples in Exercise XXII by the principles 
of right triangles. 



Then 


b 2 - x 2 


Hence 


b 2 -x 2 




.-. X 



144 PLANE TRIGONOMETRY 

20. A tree stands at a distance from a straight road and between two 
milestones. At one milestone the line to the tree is observed to make 
an angle of 25° 15' with the road, and at the other an angle of 45° 17'. 
Find the distance of the tree from the road. 

21. From the decks of two ships at C and D, 880 yd. apart, a cloud A, 
in the same vertical plane as C and B and between them, is observed. Its 
angle of elevation at C is found to be 35°, and at B 64°. Find the height 
of the cloud above the surface of the sea, the height of the eye in each 
case being 21 ft. 

22. To determine the distance between two ships at sea, an observer 
noted the interval between the flash and report of a gun fired on board 
each ship, and measured the angle which the two ships subtended. The 
intervals were 4 seconds and 6 seconds respectively, and the angle 48° 42'. 
Find the distance between the ships, the velocity of sound being 1142 ft. 
per second. 

23. In order to find the breadth of a river a base line of 500 yd. was 
measured in a straight line close to one side of it, and at each extremity 
of the base the angle subtended by the other end and a tree upon the 
opposite bank was measured. These angles were 53° and 79° 12' respec- 
tively. Find the breadth of the river. 

24. A straight road leads from a town A to a town B, 12 mi. distant ; 
another road, making an angle of 77° with the first, goes from A to a 
town C, 7 mi. distant. Find the distance between the towns B and C 

25. Two lighthouses A and B are 11 mi. apart. A ship C is observed 
from them to make the angles BAC = 31° 13 / 31" and ABC = 21° 46' 8". 
Find the distance of the ship from A. 

26. Two posts A and B are separated by a swamp. To find the dis- 
tance between them a point C is so taken that both posts are visible from 
it. By measurement, AC = 1272.5 ft., BC = 2012.4 ft., and ZACB 
= 41° 9" ll". Find the distance AB. 

27. Two buoys A and B are one half mile apart. Find the distance 
from A to a point C on the shore if the angles ABC and BAC are 77° 7' 
and 67° 17' respectively. 

28. The elevation of the top of a spire at one station, A, was 23° 50' 15", 
and the horizontal angle at this station between the spire and another 



PROBLEMS 145 

station, B, was 93° 4' 15". The horizontal angle at B was 54° 28' 30", 
and the distance between the stations 416 ft. Find the height of the 
spire. 

29. In order to find the distance of a battery at B from a fort at F, 
distances BA and AC were measured to points A and C from which 
both the fort and the battery were visible, the former distance being 
2000 and the latter 3000 yd. The following angles were then measured : 
Z BAF = 34° 10', ZFAC = 74° 42', and Z FCA = 80° 10'. Find the 
distance of the fort from the battery. 

30. The distances of two islands from a buoy are 3 and 4 mi. respec- 
tively. The islands are 2 mi. apart. Find the angle subtended by the 
islands at the buoy. 

31. Two rocks in a bay are c yd. apart, and from the top of a cliff in 
the same vertical plane with the rocks their respective angles of depression 
are A and 3 A. Show that the height of the cliff is c sin 3 A / (2 cos A). 

32. A person wishes to find the distance between two places A and B 
on opposite sides of a brook. He walks from B to a bridge 2 mi. away. 
Crossing this he continues his walk 6 mi. in the same direction to C, 
which he knows to be 3 mi. from A. If A is 4 mi. from the bridge, 
show that AB — 5.86 mi., nearly. 

33. A person at the top of a mountain observes the angle of depression 
of an object in the horizontal plane beneath to be 45°; turning through an 
angle of 30° he finds the depression of another object in the plane to be 
30°. Show that the distance between the objects is equal to the height of 
the mountain. 

34. From a window on a level with the bottom of a steeple the angle of 
elevation of the top of the steeple was 40°. At another window 18 ft. 
vertically above the former, the angle of elevation was 37° 30'. Find the 
height of the steeple. 

35. Find what angle a tower will subtend at a distance equal to six 
times the height of the tower. Find where an observer must station 
himself that the angle of elevation may be double the former angle. 

36. Two ships are a mile apart. The angular distance of the first ship 
from a fort on the shore, as observed from the second ship, is 35° 14' 10"; 
the angular distance of the second ship from the fort, observed from the 
first ship, is 42° ll 7 53". Find the distance in feet from each ship to 
the fort. 



146 PLANE TRIGONOMETRY 

37. The sides of a triangle are 17, 21, 28. Prove that the length of a 
line bisecting the greatest side and drawn from the vertex of the opposite 
angle is 13. 

38. Along the bank of a river is drawn a base line of 500 ft. The 
angular distance of one end of this line from an object on the opposite 
side of the river, as observed from the other end of the line, is 53° ; the 
angular distance of the second extremity from the same object, observed 
from the first extremity, is 79° 12'. Find the breadth of the river. 

39. Two forces, one of 410 lb. and the other of 320 lb., make an angle 
of 51° 37'. Find the size and direction of their resultant. 

40. An unknown force combined with one of 128 lb. produces a 
resultant of 200 lb., and this resultant makes an angle of 18° 24' with the 
known force. Find the size and direction of the unknown force. 



Areas and Regular Polygons 

41. Two sides of a parallelogram are 59.8 ch. and 37.05 ch., and the 
included angle is 72° 10'. Find the area. 

42. The three sides of a triangle are 49 ch., 50.25 ch., and 25.69 ch. 
Find the area. 

43. One side of a regular pentagon is 25. Find the area. 

44. One side of a regular decagon is 46. Find the area. 

45. In a circle with a diameter of 125 ft. find the area of a sector with 
an arc of 22°. 

46. In a circle with a diameter of 50 ft. find the area of a segment 
with an arc of 280°. 

47. A building is 37.54 ft. wide and the slope of the roof is 43° 36'. 
Find the length of the rafters. 

48. What angle at the center of a circle does a chord which is 4/7 of 
the radius subtend ? 

49. The side of a regular pentagon is 2. Find the radius of the 
inscribed circle. 

50. The side of a regular decagon is 23.41 ft. Find the radius of the 
inscribed circle. 



REGULAR POLYGONS 147 

51. The perimeter of a regular polygon of 11 sides is 23.47 ft. Find 
the radius of the circumscribed circle. 

52. The perimeter of a regular heptagon inscribed in a circle is 12. 
' Find the radius of the circle. 

53. Find the perimeter of a regular decagon circumscribed about a 
unit circle. 

54. Find the perimeter of a polygon of 11 sides inscribed in a unit 
circle. 

55. The perimeter of an equilateral triangle is 17.2 ft. Find the area 
of the inscribed circle. 



FORMULAS 



(sin A csc A = 1. 
cos A sec A = 1. 
tan ^4 cot J. = 1. 

2. tan J. = sin ^4 /cos -.,4. 

3. cot ^4 = cos A /sin A. 

4. sin 2 A + cos 2 J. = 1. 

5. tan 2 A + 1 = sec 2 J . 

6. cot 2 A + 1 = csc 2 A. 

7. sin (A + B) = sin ^4 cos 5 + cos A sin 5. 

8. cos (A + B) = cos J. cos B — sin ^4 sin B. 
tan .4 + tan B 



Page 



1 — tan A tan 5 
cot A o,ot B — 1 



9. tan (4 + B) 

10. cot M + B) - 

v y cot B + cot J. 

11. sin (A — B) = sin ^4 cos i? — cos ^4 sin j3 

12. cos (A — B) = cos ^4 cos B + sin \4 sin B. J 

tan ^4 — tan B 

13. tanM— J5)=- — -• 

v J 1 + tan A tan B 



14. cot (A - B) = 



cot ^4 cot 23 + 1 
cot B — cot ^4 
148 



31 



53 
56 
57 

55 

56 

57 



FORMULAS 

15. sin 2 A = 2 sin A cos A. 



16. 



cos 2 A = cos 2 .4 — sin 2 .! (i) 
= l-2sin 2 ^ (ii) 
ee 2 cos 2 ,4-1. (iii) 



17. tan 2,4 ee 

18. cot 2 A = 



19. sm- = yj- 



2 tan A 
1 - tan 2 A ' 

cot 2 ,4-1 
2 cot A 

— cos A ' 



20. cos 



w 



2 

+ COS y4 



21. tan -ee 



2 

.4 ,1 — cos A 

2 = \1 + cos A J 



.1 



fc l s ^ 



1 + cos .4 



cos A 



™ • • rt . C + Z> C-Z>] 

23. sm C + sin D ee 2 sm — - — cos — - — 



r.. • ^ C + D . C - D 
24. sin C — sm £> ee 2 cos — - — sm — - 



n C + D C- D 
25. cos C + cos D ee 2 cos — - — cos — 



C + D C D 

26. cos C — cos D ee — 2 sin — - — sin — - — 

sin C + sin D _ tan %(C + D) 
sin C — sin D ~ tan ^ (C — D) 



28. 



149 

Page 

57 

57 
58 



58 



59 



60 



sin ^4 sin B sin C 



61 

75 



150 



PLANE TRIGONOMETRY 



29. a 2 = b 2 + c 2 - 2 be cos A. 

b 2 + c 2 - a 2 



30. cos A = 
31 



2 be 
a + b tan \(A + B) 



a — b tan % {A — B) 

oo 4. ^ -^ ft-6 -C 
32. tan — - — = — — -cot — 

2 a + b 2 



33. sin 

34. cos 



_ Us-b)(s-c) 

-4 



2 \ be 

A \s(s — a) 



35. ten | = J(» -/>(»-«) . 

2 \ 5 (s — a) 



36. r = V(s-a)(s-6)(s-c)/s. ' 

37. tan (4/2) =/•/(«-«). 

38. F=flcsin4/2. 



Page 

75 



79 



88 



39. F = Vs (s -«)(*- 6) (s - c). 

- ^ „ # 2 sin ^4 sin C 

40. jF = . 

2 sin jS 

41. 7r radians = 180° = 2 right angles. 

42. N=s/r. 

43. sin (270° ± ,4) ee cos (180° ± J) 

ee - sin (90° ±A) = -cosA. 

44. cos (270° ±,4) = -sin (180° ±.4) 

= - cos (90° ± A) = ± sin A . 

45. tan (270° ± A) = - cot (180° ± A) 

= tan (90° ± ^4) = qp cot J. J 



89 

92 

95 
96 

► 44 



FORMULAS 



151 



46. sin 3 A = 3 sin A — 4c sin 3 ^4. 

47. cos 3.4 = 4 cos 3 ^4 — 3 cos A. 
3 tan A — tan 3 yl 



48. tan 3. 4 = 



1 -3tanM 



136 



49. In the following table any two expressions in the same 
line are equal arithmetically. Whether they are like or oppo- 
site in quality is known by § 22, when it is known in what 
quadrant A is. See pages 32, 33. 





sin .4 




tan A 


1 
^1 + cot 2 ^ 

cot A 








^sec 2 ^-l 
sec .4 

1 
sec .4 


1 




^1-cos 2 ^ 
cos .4 




^1 + tan 2 ^ 

1 


esc .4 




^csc^-l 


cos .4 


^l-sin 2 ^ 

sin .4 
v'l-sin 2 ^ 


^1 + tan 2 ^ 
tan .4 

1 
tan A 


^l + cot 2 ^ 

1 
cot A 

cot A 


csc .4 




^l-cos 2 ^ 
cos A 

cos A 


1 


tan A 


^sec^-l 

1 
^sec 2 ^-l 

sec^i 

sec A 


^csc 2 ^4-l 




sin .4 

1 




cot A 


^csc^-l 


v'l-cos 2 ^ 

1 
cos .4 

1 




^l + cot 2 .4 
cot .4 


csc A 


sec A 


v'l + tan 2 ^ 


^1- sin 2 .4 

1 

sin A 


^QSQ^A-l 




^1 + tan 2 ^ 
tan A 




esc A 


^l + cot 2 ^ 


csc A 


^1-cos 2 ^ 


v sec 2 A- 1 





50. If sin 6 = sin .4, 6 = nir + (- l) n A. 

51. If cos 6 = cos A, = 2 nir ± A. 

52. If tan 6 = tan A, 6 = nir + A. 

m ± n 



53. tan -1 /^ ± tan _1 « = tan -1 ; 



Page 

99 
100 
101 

107 



=F 



ANSWERS 



EXERCISE I 



1. sin A = 2/5, esc .4 = 5/2, cos ^4 = V21/5, sec A = 5/V21, 
tan .4 = 2/ V21, cot ^4 = V21 /2 ; 4 = 23° 34' 40". 

2. sin ^4 = 4/5, esc ^1 = 5/4, cos ^4 = 3/5, sec A = 5/3, 
tan -4 =4/3, cot J. = 3/4; A = 53° 8'. 

3. sin ^4 = V7/4, csc^4 = 4/V7, cos J. = 3/4, sec^4 = 4/3, 
tan ^4 =V7/3, cot ^4 = 3/V7 ; ^ = 41° 24' 30". 

4. sin .4 = V8/3, esc J. = 3/V8, cos J. = 1/3, sec A = 3, 
tan ^4 = V8, cot A = 1/ V8 ; ^4 = 70° 32'. 

5. sin A = 1/V17, esc ^4 = V17, cos ^4 = 4/VH, sec ^4 = V17/4, 
tan J. = 1/4, cot ^4 =4; ^4 = 14° 2' 15". 

6. sin ^4 = 4/5, esc ^4 = 5/4, cos ^4 = 3/5, sec A = 5/3, 
tan A = 4/3, cot J. = 3/4; A = 53° 7' 45". 

7. sin ^4 = 2/V29, esc ^4 = %/29/2, cos ^4 = 5/V29, sec A = V29/5, 
tan -4. = 2/5, cot ^4 = 5/2; ^4 = 21°48 / 6 // . 

8. sin A =3/V10, esc ^4 = V10/3, cos .4 = 1/V10, sec A = ylO, 
tan A = 3, cot ^4 = 1 /3 ; ^4 = 71° 34'. 

9. sin ^4 = 4/5, esc A = 5/4, cos A = 3/5, sec A = 5/3, 
tan A =4/3, cot J. =3/4; A = 53° 8'. 

10. sin J. = V7/4, esc A = 4/V", cos J. = 3/4, sec ^4 = 4/3, 
tan^4=V7/3, cot A = 3/v'7 ; ^4 = 41° 24' 30". 

11. sin J. = 2/5, esc A = 5/2, cos A = V21/5, sec A = 5/V21, 
tan ^4 = 2/V21, cot A = V21/2; ^4 = 23° 35". 

12. sin ^4 = 2/3, esc A- 3/2, cos A = V5/3, sec ^4 = S/V5, 
tan ^ = 2/V5, cot A = V5/2 ; ^4 = 41° 48' 30". 

163 



154 PLANE TRIGONOMETRY 

13. sin A = 4/V17, esc A = yl7/4, cos ^4 = 1/V17, sec ^4 =V17, 
tan .4 = 4, cot A = 1 /4 ; A = 75° 57 / 49". 

14. sin A = 1 /V50, esc J. = V50, cos J. = 7 / V^O, sec A = V50/7, 
tan A = 1/7, cot ^4 = 7 ; ^4 = 8° V 48". 

15. sin ^4 = 9/V82, esc ^4 = V82/9, cos ^4 = 1/V82, sec J. = V82, 
tan A = 9, cot A = 1 /9 ; 4 = 83° 39' 35". 

16-18. For answers see table on page 151. 



EXERCISE II 

1. sin 60°; cos 30°; tan 55°; cot 75°; esc 5°; sec 14°; cos 16°^ 
sin 24° 17'. 



2. 


45°. 


4. 


18°. 6. 


5°. 


8. 90° /(m + n), 


3. 


30°. 


5. 


36°. 7. 
EXERCISE 


5°. 
Ill 


9. 60°/(c-l). 


1. 


B = 65°, 




6 = '64.335, 




c = 70.98. 


2. 


A = 35°, 




a = 14.281, 




c = 17.434. 


3. 


B = 25°, 




a = 63.441, 




6 = 29.582. 


4. 


A = 75°, 




a = 74.64, 




c = 77.28. 


5. 


A = 55°, 




6 = 35.01, 




c = 61.05. 


6. 


A = 35°, 




6 = 49.152, 




a = 34.416. 


r. 


A = 20°, 




B = 70°, 




c = 106.4. 


8. 


A = 25°, 




£ = 65°, 




c = 55.15. 


9. 


^ = 20°, 




B = 70°, 




a = 34.2. 


10. 


^4 = 20°, 




5 = 70°, 




6 = 46.985. 


11. 


A = 15°, 




a = 10.352, 




6 = 38.636. 


12. 


5 = 80°, 




a = 5.289, 


• 


c = 30.45. 


13. 


5 = 70°, 




a = 27.36, 




6 = 75.176. 









ANSWERS 


155 


14. 


A = 65°, 




b = 13.989, 




c = 33.09. 


15. 


A = 65°, 




B = 25°, 




b = 14.086. 


16. 


A = 15°, 




B = 75°, 
EXERCISE 


IV 


c = 51.75. 


1. 


160.7 yd. 


5. 


42.68 ft. 




9. 81.98 ft. 


2. 


50°. 


6. 


107.22 ft. 




11. 1174.6 ft. 


3. 


96.06 ft. 


7. 


136.63 ft. 




12. 3770 ft, 


4. 


122.02 ft. 


8. 


44.78 ft., 37.58 ft. 


13. 98.097 ft,, 68.69 ft 



14. 36.08 ft., 154.23 ft. 

15. 33.51 ft., 28.12 ft. to nearest tower. 

16. 74.24 ft., 51.98 ft. 17. 378.21 ft., 417.17 ft. 

18. h = altitude = 32.14 ft., c = base = 76.6 ft., 
Q = area = 1231 sq. ft. 

19. ZC = 40°, Z A = 70°, h = 93.97 ft., Q = 3213.8 sq. ft. 

20. 61.04ft, Z^ = 35°, ZC=110°. 

21. h = 62.836 ft,, r = 76.71 ft., Q = 2764.8 sq. ft. 

22. h = 107.23 ft., r = 118.3, Q = 5361.3 sq. ft. 

23. c = r = 57.74ft., Q = 1443. 5 sq.ft. 28. 492.4ft. 

24. 20°. 29. 2515 ft. 

25. R = 274.75ft., area = 237150sq. ft. 30. 125(V3 + 3)ft. 

26. 25°; 14.09 ft. 31. 30°. 

27. 917.136 ft. 

EXERCISE V 

1. 2dqdt. 4. 2dqdt.; 4th qdt. 7. 2dqdt.; 1st qdt. 

2. 4th qdt. 5. 2d qdt.; 1st qdt. 8. 4th qdt. ; 3d qdt. 

3. 2d qdt. 6. 4th qdt. ; 3d qdt. 9. 3d qdt. ; 3d qdt. 



156 PLANE TRIGONOMETRY 

10. 847° is coterminal with 127°; 1111° with 31°; -225° with 135°; 

- 300° with 60° ; 942° with 222° ; - 1174° with - 94° or 266°. 

11. 405° and 1125° ; - 315° and - 675° ; 390° and 750° ; - 330° and 

- 690° ; 460° and 820° ; - 260° and - 620° ; 560° and 1280° ; 

- 160° and - 520°; 350° and 710°; - 370° and - 730°; 260° 
and 620° ; - 460° and - 820°. 

12. -75°; 15°. 15. - 224°22 / 17"; -134°22' 17". 

13. -138°; -48°. 16. 122°14 / 21 // ; 212° 14' 21". 

14. -205°17 / 14 // ; -115°17 / 14 // . 17. 255° 28' 42"; 345° 28' 42". 
18. 60°. 19. 175°. 20. 30°. 21. 340°. 22. 317°. 

EXERCISE VI 

1. MP = -2, OP=3, OM=tV5; osc.i = -8Acosi = TVfiA 
sec 4 ==F 3/vo, tan J. =± 2/V&, cot A = ± V5/2. 

2. MP=± 5, OM = ± 2, OP = V29; cot .4 = 2/5, 

sin A =±5/V29, esc A =± V29/5, cos 4 = ± 2/V29, 
sec A =±V29/2. 

3. MP = ± 3, OM = T 1, OP = V10 ; cot A = - 1/3, 
sin A =±3/V10, esc A =±VM)/3, cos 4 = =F 1/V10, 
sec A = =F V10. 

4. Oitf = 2, OP = 3, MP = ± V5 ; sec 4 = 3/2, sin A = ± V5/3, 
esc A =±3/V5> tan ^4 = ±vV2, cot J. =±2/V5. 

5. esc ^4 =- 8/7, cos 4 =T V15/8, sec J. =T 8/V15, 
tan J. =± 7/V15, cot A = ± V15/7. 

6. cot^4 = l/7, sin^L=±7/(5V2), esc A = ±5V2/7, 
cos A = ± 1 /(5 V2), sec ^ = ± 5 V2. 

7. sec 4 =-7/3, sin A =±2 V10/7, esc 4 = ± 7/(2 VH>), 
tan 4 =T2 V10/3, cot J. = T 3/(2 V10). 

8. ifP = ±3, Oif=±5, OP = V34; tan 4 = 3/5, sin 4 =.±3/^34, 
cscJ. =± V34/3, cos 4 = ± 5/V34, sec 4 =± V34/5. 

9. sec./! =- 5/4, sin A = ± 3/5, esc A = ± 5/3, tan ^4 = =F 3/4, 
cot J. ==F 4/3. 

10. OM = 1, OP = 2, ifP = ± V3 ; cos 4 = 1/2, sin A=± y/%/% 

esc A =± 2/V3, tan J. = ± V3, cotA=± 1/V 3 - 



ANSWERS 157 

11. cos ^4 =- 2/3, sin ^4 = ± V5/3, esc A = ± 3/V&» 
tan ^4 ==F Vo/2, cot ^4 = =F 2/V& 

12. sin ^4 =— 3/5, cos J. = T 4/5, sec ^4 = =f 5/4, tan J. = ± 3/4, 
cot ^4 — ± 4/3. 

EXERCISE VH 

1. esc A = - 3/2, cos A = T Vl -4/9 = T V^/3, sec ^4 = =F 3/ V5, 
tan ^4 = ± 2/V5, cot ^4 = ± V5/2. 

2. sec ^4 = 3, sin A = ± 2 V2/3, esc A = ± 3/(2 V2), tan ^4 = ± 2 V2, 
cot ^4 =±1/(2 V2). 

3. esc ^4 = 5, cos ^4 =± 2 V6/5, sec ^4 = ± 5/(2 V^), 
tan 4 = ± 1 /(2 V6), cot ^4 = ± 2 V6. 

4. sec J. =— 4/3, sin .4 = ± V7/4, esc ^4 = ± 4/V7, 
tan ^4 =^V7/3, cot v4 ==F 3/V7. 

5. cot J. = — 3/4, sin ^4. = ± 4/5, esc w4 = ± 5/4, cos A = =f 3/5, 
sec J. ==F 5/ 3. 

6. tan A = - 1/2, sin A = ± 1/V5, esc A = ± V5, cos^4 ==F 2/V5, 

sec A = =f V&/2. 

7. tan ^4 = 2/3, sin yt = ±2/V13, esc J. = ±V13/2, 
cos .4. =±3/V13, sec ^4 = -bV13/3. 

8. cot A = 2/5, sin ^4 =± 5/V29, esc A =± V29/5, 
cos A =±2/V29, sec ^4 =± V29/2. 

9. sin A = - 1/V3, cos J. = q=V(2/3), sec ^4 ==F V(3/2), 
tan J. = ± 1/V2, cot J. = ± y/2. 

10. cos^4 = 1/4, sin^4 = ± V15/4, esc J. = ±4/V15, tan^4 = ± V*5, 
cot ^4 =± 1/V15. 

11. cot J. =- 1/V7, sin A =±V14/4, esc ^4 =±4/V14, 

cos J. =±V2/4, sec^4 ==F 4/V2. 

c . ^ Vc 2 — m 2 . c 

12. sec A = — , sin ^4 = ± , esc A = ± - 



V^2~ 



Vc 2 - m 2 
tan A =± , cot A = ± 



m Vc 2 - m 2 

13-17. See table on page 151. 



158 PLANE TRIGONOMETRY 

EXERCISE IX 

1. sin 12°. 5. -cos 25°. 9. cos 30°. 13. esc 36°. 

2. -tan 43°. 6. sin 6°. 10. cot 25°. 14. -tan 16° 54'. 

3. sin 17°. 7. -cot 24°. 11. -cot 26°. 15. -tan 33° 39'. 

4. cos 24°. 8. sin 22°. 12. -esc 23°. 

EXERCISE X 

1. The cosine of the sum\ _ f cos first • cos second 

of any two angles J ~~ \ — sin first ■ sin second. 

2. (V2+V6)/4. 6. 1; 0. 

3. (V6-V2)/4. 7. 0; 1. 

4. (V6-V2)/4; (V6 + V2)/4. 8. A(l ,+ V42); T V(V21 -4 \/2). 

5. (V6-V2)/4; (V6 + V2)/4. 9. ^(8 + 5V3); A (2 V5 - V15). 

EXERCISE XI 

1. The cosine of the difference"! _ J cos first ■ cos second 

of any two angles J ~l + sin j^rsi • sin second. 

2. (V6-V2)/4; (V6 + V2)/4. 3. (V6-V2)/4; (V6 + V2)/4. 

4. (2V2-V15)/12; (2V30 + 1)/12. 

5. (3V5-2 V7)/12; (6 -f V35)/12. 

EXERCISE XII 

1. The tangent of the difference \ _ f tne difference of their tangents 

of any two angles J ~~ \ 1 + product of their tangents 

2. (V3 + 1)/(V3-1). 4. 1; 7. 

3. (V3-1)/(V3 + 1). 5. 1; 1/7. 

n _ cot J? 4- cot ^L cot 5 -cot ^4 

12. tan ( A 4- -S) = 5 tan ( J. - B) = ■ 



" cot .A • cot I? — 1 cot ^4 • cot B + 1 

j. / a . ™ 1 — tan A • tan i? . - . . _. 1 + tan A • tan B 

13. cot (A + #) = ; cot M. - B) = 

x tan J. + tan J5 tan J. - tan i? 



ANSWERS 159 



EXERCISE Xni 

^ __ " . . twice the tangent of the angle 

1. The tangent of twice an angle = — . 

1 — (the tangent of the angle)' 2 

2. V3/2; 1/2; V3. 

3. V3/2; -1/2; - V3. 

4. 2 sin 34 cos 3 4 ; cos 2 3 4- sin 2 3 4, 1 - 2 sin 2 3 4, or 
2 cos 2 3 A - 1 ; 2 tan 3 -4 / (1 - tan 2 3-4). 

5. 2 sin (3 A/2) cos (3 A/2); cos 2 (3 4/ 2) - sin 2 (3 A/2), or 

1 -2 sin 2 (3 4/2), or 2 cos 2 (3 4/2) - 1; 

2 tan (3 4 /2) / [1 - tan 2 (3 4/2)]. 



EXERCISE XIV 

1 + cos angle 



1. cos half an angle = square root of 
tan half an angle = square root of 



2 
1 — cos angle 
1 + cos angle 



V2 



2. V2-V2/2; V2+V2/2; A /- — = V3 - 2 V2. 

\ 2 + V2 



3. V2 - V3/2; V2 -+- V3/2 ; V? - 4 V3. 

4. V3/3; V6/3; V2/2. 

fl — a ll + fl II -a 

\"T" ; \2~ ; \lTa' 

/l-cos24 # /l + cos24 > ll-cos2 4 

\ 2~~ ' \ 2 ' \1 + cos24' 

II— cos 4 4 II + cos 44 # II — cos 4 4 

\ 2 ; \ 2 ' \1 + cos44* 

/l — cos 6 4 /l + cos64 > /l — cos 6 4 

' \ <T~ ; \ 2~~ ; \l + cos64' 



160 



PLANE TRIGONOMETRY 



EXERCISE XV 



f twice cos half sum into 
^ sin half difference. 



The difference of the sines of "1 
any two angles j 

The sum of the cosines of any 1 
two angles J 

The difference of the cosines 1 f twice sin half sum into 

of any two angles ) \ sin half difference. 



J twice cos half sum into 
\ cos half difference. 



15. (1) sin (A + B) = (2 V2 + V3)/6 ; sin (J. - B) = (2 V2 - V3)/6 ; 
cos (J. +■ J5) = (2 V6 - I)/ 6 ; cos (^ - B ) = ( 2 V6 + l)/6 ; 
sin 2^1 =V3/2; sin 2 J5 = 4 V2/9; 

cos 2^1 = 1/2; cos 2 5 = 7/9; 



(2) sin (A + B) = - (2 V2 - V3)/6 ; 
cos (J. + £) = - (2 V6 -f l)/6 ; 
sin2^L= V3/2; 
cos 2 ^4 = 1/2; 



sin (A-B) = -(2y/2 + V3)/6 ; 
cos (4 - £) = - (2 V6 - l)/6 ; 

sin 2£ = -4 V2/9; 

cos 2 5 = 7/9. 



16. (1) 



(2) 



tan (J. + B) = 

tan (4 - B) = 

cot (ud + 5) = 

cot (4 - B) -. 

sec (^4 + B) - 

esc (^. + B) z 

tan 2 ^4 : 

cot 2 ^L : 

sec 2 5 : 

esc 2 B z 

tan (^4 + B) : 
tan (4 - B) : 

cot (i + 5): 

cot (A - B)z 

sec (^4 -f 5) : 
esc (A -{- B)z 

tan 2 A : 

COt 2 ^4 : 

sec 2 jB : 

CSC 2 £ : 



;(2V2+V3)/(2V6-1) 

:(2 V2-V3)/(2 V6 + 1) 

:(2V6-1)/(2V2+V3) 

:(2V6 + 1)/(2V2-V3) 

:6/(2V6-l); 

:6/(2V2 + V3); 

V3; 

:1/V3; 

:9/7; 

:9/(4V2). 

:(2V2-V3)/(2V6 + 1) 

:(2V2+V3)/(2V6-1) 

:(2V6 + 1)/(2V2-V3) 

:(2V6-1)/(2V2+V3) 

:-6/(2V6 + l); 

:-6/(2V2-V3); 

V3; 

:1/V3; 

= 0/7; 

: _ 9/4 V2. 



ANSWERS 161 



EXERCISE XVI 



9. sin (3 x/4) = ± V[l - cos (3 x/2)] /2 ; 

cos (3 x/4) = ± V[l + cos(3x/2)]/2 ; 



tan (3 x/4) ee± Vl - cos(3x/2)/ Vl + cos(3x/2). 

10. sin (3x/4)=2sin(3x/8)cos(3x/8); 
cos (3x/4) = cos 2 (3 x/ 8) - sin 2 (3x/8) ; 
tan(3x/4) = 2tan(3x/8)/[l - tan 2 (3x/8)]. 



16. 


(1) 


sin (A + B) = 
cos (A -f B) = 

sin 2 A - 
sin 2 B = 


(2 +V15)/6; sin (A - 
:(V5-2V3)/6; cos(A- 
:4vV9; cos 
;V3/2; cos 


-J3) = (2-V15)/6; 
-£) = (V5 + 2V3)/6 
2^4 = 1/9; 
2B = - 1/2. 




(2) 


tan (^4 + B) = 

tan (^4 - B) = 

tan 2 A = 
tan 2 5 = 


2+V15 , . ^ V5-2V3 
v ; cot (A + B) = w ; 

V5-2V3 2+V15 

2-V15 ,. ^ V5 + 2V3 
: ; cot (A - B) - — — - — — ; 

V5 + 2V3 2-V15 
:4 V5; cot 2 ^4 = V-^/20; 
- V3; cot2£ = - V3/3. 








EXERCISE XVn 




1. 


A 


= 23°, 


6 = 11.779, 


c= 12.796. 


2. 


B 


= 52°, 


6 = 10.355, 


c = 13.14. 


3. 


B 


= 75°, 


6 = 6.7614, 


a = 1.8117. 


4. 


A 


= 40°, 


a = 16.782, 


c = 26.108. 


5. 


A 


= 34° 22' 9", 


B = 55° 37' 51", 


6 = 0.51176. 


6. 


A 


= 33° 8' 56", 


B = 56° 51' 4", 


c = 499.26. 


7. 


A 


= 39° 49' 22", 


B = 50° 10' 38", 


a = 48.863. 


8. 


B 


= 81°, 


a = 148.41, 


c = 948.68. 


9. 


A 


= 49° 53' 53", 


5 = 40° 6' 7', 


c = 4.4632. 


10. 


B 


= 43° 37', 


a = 3821.5, 


6 = 3641.3. 


11. 


A 


= 35° 53' 56".; 


3, 5 = 54° 6' 3". 5, 


6 = 731.23. 


12. 


A 


= 66° 51', 


a = 176.53, 


c = 191.99. 



162 PLANE TRIGONOMETRY 

13. A = 71° 22', a = 2.4099, 6 = .81258. 

14. J5 = 58°15', . 6 = 77.632, c = 91.294. 

15. A = 32° 10' 15", B = 57° 49' 45", a = 388.45. 

16. .4 = 7° 53' 42", 6 = 644.11, c = 650.27. 

EXERCISE XVIII 

1. C=1$0°-2A, r = c/(2'CosA), ft = c-tan^i/2. 

2. A = 90° - 0/2, r = A/cos (C/2), c = 2 A • tan (C/2). 

3. A = t2LTi-i(2h/c), = 2.tan-i(c/2/i), r = V4/1 2 + c 2 /2. 

4. r = 2.055, h = 1.6853, -4 = 55° 5' 30", Q = 1.9819. 

5. r = 7.706, c = 3.6676, C = 27° 32', Q = 13.7253. 

6. Let x = length of rafter and Q = area of roof ; then, since the 
roof projects one foot over the side of the barn, we have : 

x = 21 /sin 45°, Q = 2 ■ x • 82 ; x = 29.698, Q = 4870.44. 

7. r = 1.61804, A = 1.53882, F = 7.69417. 

8. r = 11.2692, h = 10.8852, .F = 380.99. 

9. r = 1.0824, c = 8284, F = 3.3136. 

10. r = 1.5994, A = 1.441, p = 9.715. 

11. ft = 28.971, r = 31.357, A — (difference between polygon and in- 
scribed circle) = 144.51, D c = 307.8. 

12. h = 14.536, r = 16.134, D t = 48.48, D c = 105.47. 
15. 99.64 sq. ft. 16. 2.393. 

EXERCISE XIX 

1. A = 65° 15', 6 = 95.6025, c = 89.648. 

2. J5 = 72°14 / , a = 75.132, c = 92.788. 

3. A = 31° 20', 6 = 184.896, c = 191.978. 

4. C = 60°, a = 255.38, 6 = 282.56. 

5. ^1 = 15° 43', 6 = 222.1, c = 321.08. 



ANSWERS 



163 



6. B = 66°, 

7. 1253.2 ft. 

8. 1116.6 ft. 



a = 765.43, 
9. 300 yd. 
10. 12296 ft., 13055 ft. 



c = 1035.4. 

11. 294.77 ft. 

12. 4211.8 ft. 



1. 


A = 32° 25' 36" 


2. 


A = 28° 20' 48" 


3. 


B = 32° 36' 9", 


4. 


#i = 51° 18' 22" 




£ 2 = 128° 41' 38' 


5. 


#i = 31° 57' 46" 




£ 2 = 148° 2' 14" 


6. 


Impossible. 


7. 


Ci = 46° 18' 40" 



C 2 = 133° 41' 20 

8. ^i = 51° 18' 27 
A 2 = 128° 41' 33 

9. A = 54° 31' 13' 

10. B x = 24° 57' 54 A 
£ 2 = 155° 2' 6", 

11. £i = 16° 43' 13' 
J5 2 = 163° 16' 47 



EXERCISE XX 

C = 106° 24' 24", 

C = 39° 35' 12", 

C = 83° 33' 5", 

Ci = 88° 41' 38", 
C 2 = 11° 18 / 22", 

^x = 120° 44' 14", 
A 2 = 4° 39' 46", 

A t = 93° 9' 13", 
A 2 = 5° 46' 33", 

Ci = 98° 21' 33", 

C 2 = 20° 58' 27", 

C = 47° 44' 7", 

Ci = 133° 47' 41", 
C 2 = 3° 43' 29", 

^! = 147° 27' 47", 
A 2 = 0° 54' 13", 



c = 259.4. 
a = 293.56. 
c = 6.621. 

d = 218.525; 

c 2 = 42.853. 

ai = 120.31 ; 
a 2 = 11.379. 

a x = 69.4567 ; 
a 2 = 7.0005. 

Ci = 43.098 ; 
c 2 = 15.593. 

c = 50.481. 

Ci =615.67; 
c 2 = 55.41. 

ai = 35.519; 
a 2 = 1.0415. 



1. A = 42° 50' 58", 

2. 5 = 132° 18' 28", 

3. A = 109° 15' 30", 

4. A = 60° 44' 39", 

5. J5 = 54° r 13", 

6. 902.94 ft. 

7. 13.27 mi. 



EXERCISE XXI 

B = 64° 9' 2", 
C = 14° 34' 23", 
5 = 45° 4' 30", 
5 = 47° 21' 21", 
C = 63° 49' 9", 

8. 1331.2 ft. 

9. 4.8112 mi. 



c = 374.05. 
a = 67.75. 
c = 440.45. 
c = 965.28. 
a = 44.825. 

10. 10.532 mi. 

11. 9.6268 mi. 



164 PLANE TRIGONOMETRY 







EXERCISE XXII 




1. 


A = 74° 40' 18", 


B = 47° 46' 38", 


C = 57° 33' 4". 


2. 


A = 59° 19' 14", 


£ = 68° 34' 8", 


C = 52° 6' 40". 


3. 


^4 = 45° ll 7 50", 


5 = 101° 22' 12", 


C = 33° 25' 58". 


4. 


A = 71° 33' 52", 


jB = 49° 8' 3", 


C = 59° 18' 5". 


5. 


^4 = 54° 3 / 10", 


J5 = 30° 47' 22", 


C = 95° 9' 24". 


6. 


J5 X = 41° 41 / 26", 


d = 111° 52' 34", 


ci = 177.2 ; 




J5 2 = 138° 18' 34", 


C 2 = 15° 15' 26", 


c 2 = 50.248. 


7. 


A = 17° 16' 11", 


5 = 28° 43' 49", 


c = 14.424. 


8. 


A = 18° 12' 22", 


£ = 135° 50' 46", 


C = 25° 56' 52". 



9. Since c>a, C>A> 90°, which is impossible, as a triangle cannot 
have two obtuse angles. 

10. B = 48° 34' 44", A = 49° 38' 16", a = 76.015. 

11. B = 145° 35' 24", C = 7° 11' 36", a = 104.57. 

12. A = 57° 52' 44", B = 70° 17' 24", C = 51° 49' 50". 

13. ^Li = 70° 12' 48", B 1 = 57° 22' 56", b x = 28.79 ; 
A 2 = 109° 47' 12", B 2 = 17° 48' 32", b 2 = 10.454. 

14. A = 32° 44' 40", C = 63° 48' 28", b = 137.39. 

15. 6 = 143.52, B = 146° 43' 10", C = 14° 4' 7". 

16. ^4 = 67° 55' 15", 5 = 54° 4' 45", c = 85.36. 

17. 44° 2' 9", 51° 28' 11", 84° 29' 40". 

18. 44° 2' 56". 20. N. 4° 23' 2" W., or S. 4° 23' 2" W. 

19. 60° 51' 8". 21. 60°. 

EXERCISE XXIII 

2. 1931.8. 3. 44770; 781.617; 149689; 314543. 

4. 123794 ; 53596.3 ; 10.6665 ; F t = 11981.7, F 2 = 2349.63. 

5. 1016.23; 30.858; 1430.3; 170346. 

6. 3891.64; 3319.38; 9229.4; 31246.4. 



ANSWERS 165 

EXERCISE XXIV 

1. 9tt/4, 17 7T/4, -7tt/4, -Iott/4; 13 tt/4, 21 % / 4, -3 tt/4, 
-11 tt/4; 7 tt/2, IItt/2, -7T/2, -5 tt/2; 9 tt/2, 13 tt/2, -3 tt/2, 
-7 tt/2; 7 tt/3, 13 tt/3, - 5 ?r/3, -Htt/3; 8 tt/3, 14 tt/3, — 4 tt/3, 
-IOtt/3; 13tt/6, 25?r/6, -11tt/6, -23tt/6; 17tt/6, 29tt/6, 
- 7 7T/6, - 19tt/6. 

2. 120°. 4. 900°. 6. 7T/4. 8. tt/2. 

3. 300°. 5. 135°. 7. 3;r/4. 9. 3 tt/2. 

10. 12859/7560, or 1.7. 14. tt/4, 3 tt/4. 

11. 115643/37800, or 3.06. 15. - *r/6, *r/3. 

12. 35827/47250, or 0.76. 16. - *r/4, tt/4. 

13. 1516273/2268000, or 0.67. 17. - 7 tt/6, - 2 tt/3. 

18. sin(7T/6) = l/2, cos (?r/6) = V3/2, tan (* / 6) = V3/ 3. 

19. sin (tt/4) = V2/2, cos (* / 4) = y/2/ 2, tan (tt/4) = 1. 

20. sin (tt/3) = V3/2, cos(tt/3) = 1/2, tan(*r/3) = y/S. 

21. sin (tt/2) = 1, cos (tt/2) = 0, tan (tt/2) = go. 

22. sin re = 0, cos it — — 1, tan 7t = 0. 

23. 157/70 or 2.24; 157 (57° 17' 44 // .8)/70. 

24. 55/98; 55(57°17 / 44 // .8)/08. 27. 24 T 6 T in. 

25. 3/4 ; 3 (57° 17' 44 /7 .8)/4. 28. 68 T « T in. 

26. 11/210 in. 29. 3980^ mi. 

EXERCISE XXV 

1. $ = U7t + (-l) n (± TT/2) -UK ± Tt/2. 

2. n7t±7t/4:. 6. ?i7r± tt/6. 10. 2n7t±7t/3. 

3. ?27r±7r/4. 7. 2?i;r ± 2 7T/3. 11. 2?i7T + 7r/4. 

4. nit ± 7t/6, 8. ?i7T + (~l) n • 7t/6. 12. me ± ?r/6. 

5. n7r±7r/3. 9. mt + (-1)" • (tt/3). 13. titt + ;r/4. 



166 PLANE TRIGONOMETRY 

14. rat + tf/4, rift + 7t/S. 16. rt7t±7t/4t. 

15. ntf + 5 ?r/6, ri7r + 2 ;r/3. 17. n • 180° + 53° V 45". 

18. (2 n + 1) • 180°, 2 n ■ 180° + 53° 8'. 

19. [rwr + (-l)»*r/4]/5. 22. (2nar + *r/2)/5, : 2 7wr- ar/2. 

20. 2ft7r/9, 2717T. 23. (2n?r ± 7T/2)/(md=r). 

21. (ntf-f 7t/2)/(r + 1). 24. 2n^r + ar/4, 2wtt - 3tt/4. 
must be in the first or third quadrant. Why ? 

25. 2ft7T-7r/6, 2ft7r+5 7T/6, Init-it/Z, 2rut + 27t/3. 
6 must be in the second or fourth quadrant. Why ? 

26. 2nit- Tt/l, 2n7t + 37t/4 28. (2n + 1) it + tt/4. 

27. n,7zr/3 + 7T/6. 29. 2n^-^r/6. 

30. 7 tt/ 12 and tzt/4; (2 m + n) tt/2 ± 7t / 6 + (- l)»(jr/12) and 
(n — 2 m) 7t / 2 -f (— \) n 7t / 12 =f 7T/6, where m and n are any integers. 

31. 25 ?r/ 24 and 19tt/24; (ra + 2 n) ?r/2 + tt/8 ± tt/12 and 
(2n-m)^/2 ± tf/12 — rt/8, where m and n are any integers. 

EXERCISE XXVI 

1. rwr/4, (2n^r± tt/3)/3. 8. n*r ± ?r/4, 2rwr ±2;r/3. 

2. 717T ± 7T/4, 2W7T ± 7T/3. 9. 2 7l7t + 7t / 6 ± 7t / 4. 

3. ntf ± ?r/4, mt —(— l) n 7t/6. 10. 2n^r + ^r/4. 

4. ftTT/2 ± 7zr/8, 2wtt/3 ± nr/9. 11. 2n^ - tt/3 ± tt/4. 

5. 2n^/3±^r/6, n*r + (- l) w ar/6. 12. 2n^r + tt/4 d= tt/5. 

6. twr/3, (2w + l/3)*r/4. 13. 2 a • 180° + 68° 12'. 

7. nit 1 3, riTT ± it / 3. 14. 2 n • 180° ± 60° - 33° 41' 24". 

EXERCISE XXVII 

1. (2-V3)/2; (2-V2)/2; 1/2; (2+V2)/2; 0; 1; 2; 1. 

2. 1/2; (2-V2)/2; (2-V3)/2; (2-V2)/2; 1; 0; 1; 2. - 



ANSWERS 167 

3. n7r+(-l) n -7r/4; mt -(-1)*- rt/3; 2nn±n/Q\ 2n7t±2 7t/3; 
n7t+7t/6; mt — it/Z\ nit — it/±\ mt + it/Z. 

13. 1/2. 15. V3. 17. V3. 19. 13. 

14. ±V2/2. 16. -8,1/4. 18. ±V21/14. 

EXERCISE XXIX 

1. (1/V2 + i/V2) • V2, or (costt/4 4- isin?r/4) • y/2. 

2. [i/V2 + i(-l/V2)]- V2, or [cos(- tt/4) + i sin (- tt/4)] • V2. 

3. [_l/V2 + i(-l/V2)]- V2, 

or [cos(-3 7r/4) + isin(-3^/4)]. V2. 

4. (_ 1/2 + z V3/2)-2, or (cos 2 tt/3 + isin 2 ar/3) • 2. 

5. (-^3/2 4-i/2) -2, or (cos 5 tt/6 + i sin 5 ar/6) • 2. 

6. [-V3/2 + *(- 1/2)] -2, or [cos(- 5 tt/6) + zsin(- 5 tt/6)] -2. 

7. [3/5 +£(-4/5)]. 5, or [cos (- 53° 8') 4- i sin(- 53° 8')] • 5. 

8. (_ 3/V13 + i • 2/V13)- V13, or (cos 146° 19' + isin 146° 19') • VI 3. 

9. [_3/5 + i(_4/5)].io, or [cos (- 126° 52') 4- i sin (-126° 52')] -10. 

10. [_V5/4 + z(-VH/4)].4, 

or [cos (- 123° 590 + *sin(- 123° 59')] • 4. 

11. cos (90° - 0) 4- i sin (90° - 0). 

12. cos (90° 4- 0) 4- i sin (90° + 0). 

EXERCISE XXX 

4. Each quality unit in example 2 is a fifth root of — 1. Each quality 
unit in example 3 is a sixth root of — 1. 

5. cos (6 4- 5 0) 4- i sin (6 4- 50). 

6. cos (5 /3 - 3 0/2) 4- i sin (5 0/3 - 3 (9/2). 

7. cos (0 + - p - y) 4- £ sin (0 + - /3 - 7). 

8. cos(— 7r) 4- isin(— 7r), or — 1. 

9. cos (4 4- 5 - Tt/2) + £ sin (4 + 5 - tt/2), 
or sin (4 + 5 0) — i cos (4 4-5 0). 



168 PLANE TRIGONOMETRY 

EXERCISE XXXI 

1. +1, (-1± iV3)/2. 2. -1, (l±V33)/2. 

3. ±1, (l±»V3)/2, (-l±»V3)/2. 

4. ±i, (V3±i)/2, (-V3±*)/2. 

5. ±2, ±i-.2. 

6. 2, 2[cos(2n7T/5) ± isin(2n^r/5)], where ft = 1 or 2. 

Here two roots are the reciprocals of two others respectively. 



7. -3, 3[cos(2n + l7r/5) + isin(27i + l7r/5)], where ft = 1, 2,3, or 4. 



8. ± [cos (4 ft - 1 tt/12) + isin(4w - In:/ 12)], where n = 0, 1, or 2. 
Here the roots are opposites in pairs. 



9. 2 [cos (12 n + Itt/18) + i sin (12 ft -f 1 n/ 18)], where n = 0, 1, or 2. 

10. ±[cos(6n-l7T/12) + i sin (6 w - 1 tt/12)] 4/2, where n = or 1. 
Here the roots are opposites in pairs. 



11. ± [cos (8 n +1 ?r/24)+ i sin (8 ft + 1 *r/24)]V2, where n = 0, 1, or 2. 
Here the roots are opposites in pairs. 

12. ^4[cos(r*r/15) + isin(nr/15)], where r = - 1, 5, 11, 17, or 23. 

13. cos(7T/5) ± £sin(7r/5), cos(3^r/5) ± zsin(37r/5). 

14. ±1, (l±V^3)/2, (-1 ±iV3)/2, ±i, cos(tt/6) ± isin(*r/6), 
cos (5tt/ 6) ± isin(5^r/6). 

15. -1, (1 ±vC^/2, ± [cos(tt/4) + isin(*r/4)], 
± [cos(3tt/4) + isin(3*r/4)]. 

16. -1, cos(ft7T/7) ± isin(ft7T/7), where n - 1, 3, or 5. 

EXERCISE XXXn 

1. n • 360° + 45°, or 2 ftTT + *r/4. 

2. ft • 360° + 132°. 4. ft • 360° - 100°. 

3. ft • 360° - 35°. 5. 2 nit + *r/6. 

6. cos^l=±V33/7, tan^=±4/V33, cot .A = ± V33/4, 
sec^i = ±7/V33, csc^L = 7/4. 



ANSWERS 169 

7. cot 4 = 2/3, sin4=±3/V13, cos4 = ±2/V13, 
sec4 = ± V13/2, csc4 = i V13/3. 

8. sin A = ± V55/8, tan4 = =F V55/3, cot4 =q= 3/V55, 
sec 4 =- 8/3, esc 4 =± 8/V55. 

9. tan 4 = — 5/7, sin 4 = ± 5/V74, cos 4 = =F 7/V"4, 
' sec 4 = =fV74/7, csc4 =± V74/5. 

10. sin4 = ± V33/7, cos4 = 4/7, tan A = ± V33/4, 
cot 4 =±4/V33, csc4 = ± 7/V33. 

11. sin A = -4/5, cos4 = =F 3/5, tan A = ± 4/3, cot A = ± 3/4, 
sec A = =f 5/3. 

12. 1st or 2d ; 1st or 3d ; 2d or 3d ; 2d or 4th ; 1st or 4th ; 3d or 4th. 

13. cos 4°. 16. - cot 40°. 19. - cos 5°. 

14. -sin 38°. 17. - sec 10°. 20. - tan 20°. 

15. tan 35°. 18. - cos 15°. 21. tan 30°. 
22-27. See table on page 151. 66. 2d or 3d. 

63. 2d or 4th. 67. rut ± it/ 2. 

64. 1st or 4th. 68. n • 180° + (- 1)» 21° 28'. 

65. 3d or 4th. 69. 2mt ± tt/2, rut + (- 1)" *r/6. 

70. nit + (- l)»7T/6, nn -(- l)»7T/2. 

71. n • 180° ± 35° 16'. 72. 2 mt ± tt/3, 2 rut ± n. 73. ?irr ± 7t/6. 

74. n • 360° ± 51° 19', n ■ 360° ± 180°. 

75. rut + (- l) n 7t/6, nn - (- \) n 7t/2. 

76. fttf ± 7r/4. 

77. n • 180° + 45°, n ■ 180° - 18° 26'. 

78. n- 180° + (- 1)»30°, ft- 180°- (- 1)»19°28'. 

79. iiTt/2 + (- 1)»7T/12, ft7r/2 - (- 1)»tt/4. 

80. war + (- l)«*/6. 82. 2?itt ± 7T/4, 2nn ± n/2. 

81. Twr -(-!)« (tt/3). 83. rut + (- l)»zr/6. 



170 PLANE TRIGONOMETRY 

84. n • 180°, n • 360° + 65° 43', n ■ 360° + 204° 18'. 

85. 2n7t ± 7t/3, 2mt ± it/2. 86. n • 360° + 105°, n ■ 360° - 15°. 

87. 2n7t ±7t/2, 2n?t, 2n?t ±2?t/S. 

00 n , , , . . , , , , - a ± Va 2 + 8 a + 8 

88. 2 nit ± the principal value of cos -1 

4 

89. (2n+ 1)tt/6. 

90. w • 180° + 22° 37', n ■ 180° + 143° 8'. 

91. ±1, ±V21/7. 93. 1. 95. V3/3, -V3/2. 

92. V3/2. 94. V3/3. 96. (m + 1)tt/3. 

97. nit ± 7t/6, nit ± n/Z. 

98. nit + (- l)»ar/6, ?i • 180° + (- 1)'* ■ 14° 29'. 

99. 2 ?i7r, 2 n7T + it, 2 nit + 7r/4, 2n?r + 5 7T/4. 

100. »7T ± 7T/3. 



101. = tan- 1 (a/b); r = Va 2 + b 2 . 



102. = tan- 1 (a/b); <$> = tan- i (c / Va 2 + & 2 ); r = Va 2 + & 2 + c 2 ). 



103. x==tan- 1 (a/6) + cos- 1 (±Va 2 _-h_6 2 /2); 
y = tan- 1 (a/ 6) - cos-^i Va 2 + 6 2 /2). 



104. a = tan-i(- 6/a) + sin-^T Va 2 + 6 2 /2); 
y = tan- 1 (a/b) - sin-^i Va 2 + 6 2 /2). 

105. cc = (6 sin J5 — a cos B)/sin(B — A); 
y = (a cos A — b sin .4) /sin (B — J.). 



,-„ „ _ a cos B - 6 sin A Va 2 + 6 2 - 2 a& sin M - B) 

106. = tan— 1 ; r = • 

b cos A + a sin 5 cos (J. — 5) 

107. x — 0, y = ; x= 7t, y = — 7t ; x — — .it, y = it. 

108. i2=TT(sin^ + cosA); F = [TT- W(sinh + cos ft) cos /i]/ sin ft. 



109. a = r vers -1 (y/r) T ^2 ry — ?/ 2 . 

110. (ae - bd) 2 = (af - cd) 2 + (ce - 6/) 2 . 



ANSWERS 171 

111. a 2 + b' 2 = c 2 + d 2 . 115. 4, 1 ± V3. 

114. -4, 2±2V3. 116. - 1, (l±V3)/2. 

117. - 1 + 2 cos 40°, -1+2 cos 160°, - 1 + 2 cos 280°. 

118. -4/3 + (2 V10/3)cos^i. where A = 39° 5' 51", 
159° 5" 51", or 279° 5' 51". 

EXERCISE XXXIH 

1. 0.43498 mi. 3. 502.46 ft. 5. 71° 33' 54". 

2. 37°36 / 30 // . 4. 49° 44' 38". 6. 56.649 ft. 

7. 260.21 ft., 3690.3 ft. 10. 238,850 mi. 

8. 235.81 yd. 11. 90,824,000 mi. 

20. 0.32149 mi. 24. 12.458 mi. 28. 278.7 ft. 

21. 942.7 yd. 25. 5.1083 mi. 29. 5422 yd. 

22. 5147.9 ft. 26. 1346.3 ft. 30. 28° bV 20". 

23. 529.5 yd. 27. 0.83732 mi. 34. 210.4 ft. 

35. 9° 28' ; 2.9152 times the height of the tower. 

36. 3121.1 ft., 3633.5 ft. 38. 529.49 ft. 

39. 658.36 lb., 22° 23' 47" with first force. 

40. 88.326 lb., 45° 37 x 16" with known force. 

41. 210 acres 9.1 sq. ch. 42. 61 acres 4.97 sq. ch. 

43. 1075.3. 47. 25.92 ft. 51. 3.7865. 

44. 16,281. 48. 33° 12" 4". 52. 1.9755. 

45. 749.95 sq.ft. 49. 1.3764. 53. 6.4984. 

46. 1834.95 sq. ft, 50. 36.024 ft, 54. 6.1981. 

55. 8.6058 sq. ft. 



AN371S04 





B 4/6/12 
275 


II 

a I a 














r 



18 20 1 22 I ! 



